Chemistry the molecular nature of matter and change 7th edition silberberg solutions manual

Chemistry The Molecular Nature of Matter and Change 7th Edition Silberberg Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/chemistry-the-molecular-nature-of-matter-and-change-7th-edition-silberberg-sol utions-manual/

CHAPTER 7 QUANTUM THEORY AND ATOMIC STRUCTURE The value for the speed of light will be 3.00x108 m/s except when more significant figures are necessary, in which cases, 2.9979x108 m/s will be used.

FOLLOW–UP PROBLEMS 7.1A

7.1B

Plan: Given the frequency of the light, use the equation c = λν to solve for wavelength.

Solution:

λ=

c 3.00x108 m/s  1 nm  =   = 414.938 = 415 nm ν 7.23x1014 s −1  10−9 m 

λ=

c 3.00x108 m/s  1 Å  =   = 4149.38 = 4150 Å ν 7.23x1014 s −1  10−10 m 

Plan: Given the wavelength of the light, use the equation c = λν to solve for frequency. Remember that wavelength must be in units of m in this equation.

Solution:

ν=

c

λ

=�

3.00x108 m/s

109 nm

940 nm

1m

��

This is infrared radiation.

7.2A

� = 3.1915x1014 = 3.2x1014 s–1

Plan: Use the formula E = hν to solve for the frequency. Then use the equation c = λν to solve for the wavelength.

Solution:

ν=

–19

8.2x10 J E � = 1.2375x1015 = 1.2x1015 s–1 =� –34 h 6.626x10 J•s

(using the unrounded number in the next calculation to avoid rounding errors) λ= 7.2B

3.00x108 m/s 109 nm c � � � = 240 nm =� 1m 1.2375x1015 s–1 ν

Plan: To calculate the energy for each wavelength we use the formula E = hc/λ.

Solution:

(

)

(

)

(

)

6.626x10−34 J • s)(3.00x108 m/s hc = = 1.9878x10–17 = 2x10–17 J λ 1x10−8 m 6.626x10−34 J • s)(3.00x108 m/s hc = = 3.9756x10–19 = 4x10–19 J E= λ 5x10−7 m 6.626x10−34 J • s)(3.00x108 m/s hc = = 1.9878x10–21 = 2x10–21 J E= −4 λ 1x10 m As the wavelength of light increases from ultraviolet to visible to infrared, the energy of the light decreases.

E=

7-1

7.3A

 1 1  − 2 Plan: Use the equation relating ΔE = – 2.18x10– 18 J  2  to find the energy change; a photon in n n initial   final the IR (infrared) region is emitted when has n has a final value of 3. Then use E = hc/λ to find the wavelength of the photon.

Solution:  1 1  − 2 a) ΔE = – 2.18x10– 18 J  2  n n initial   final

1   1 ΔE = – 2.18x10– 18 J  2 − 2  6  3 ΔE = –1.8166667x10– 19 = –1.82x10– 19 J b) E = λ =

7.3B

hc λ

(

)(

)

6.626x10−34 J • s 3.00x108 m/s  1 Å  hc 4 4 =  −10  = 1.094202x10 = 1.09x10 Å −19 E 1.8166667x10 J  10 m 

Plan: Use the equation E = hc/λ to find the energy change for this reaction. Then use the equation  1 1  − 2 ΔE = – 2.18x10– 18 J  2  to find the final energy level to which the electron moved. n n initial   final Solution: (6.626x10–34 J•s) (3.00x108 m/s)

��

109 nm

–4.85x10–19 J

6.022x1023 H atoms

1 kJ

1 H atom

1 mol H

a) ΔE = hc/λ = �

410. nm

1m

� = ΔE = 4.8483x10–19 = 4.85x10–19 J

Because the photon is emitted, energy is being given off, so the sign of ΔE should be negative. Therefore, ΔE = –4.85x10–19 J ΔE (kJ/mol) = �

��

number)

 1 1  − 2 b) ΔE = – 2.18x10– 18 J  2  n n initial   final 1 n2final 1

=

ΔE

+ J

n2initial

–19

J

1

–4.85x10

� = –292 kJ/mol (number of atoms is a positive

1000 J

1

–18

–2.18x10

��

= + 2 –18 n2final 6 –2.18x10 J 1

= 0.25025 n2final

n2final = 3.9960 = 4 n final = 2 7.4A

(The final energy level is an integer, so its square is also an integer.)

Plan: With the equation for the de Broglie wavelength, λ = h/mu and the given de Broglie wavelength, calculate the electron speed. The wavelength must be expressed in meters. Use the same formulas to calculate the speed of the golf ball. The mass of both the electron and the golf ball must be expressed in kg in their respective calculations.

7-2

Solution: a) λ = h/mu u=

h = mλ

(

 kg • m 2 /s 2  6.626 x10−34 J • s   = 7273.3 = 7.27x103 m/s −9 J     10 m   9.11x10−31 kg (100.nm )      1 nm  

)

b) Mass (kg) of the golf ball = (45.9 g) � u=

6.626x10–34 J•s h = 10–9 mλ (0.0459 kg)�(100. nm)�

7.4B

��

1 nm

�

� = 0.0459 kg

1000 g

kg•m2 /s2 J

� = 1.4436x10–25 = 1.44x10–25 m/s

Plan: Use the equation for the de Broglie wavelength, λ = h/mu with the given mass and speed to calculate the de Broglie wavelength of the racquetball. The mass of the racquetball must be expressed in kg, and the speed must be expressed in m/s in the equation for the de Broglie wavelength.

Solution: Mass (kg) of the racquetball = (39.7 g) � Speed (m/s) of the racquetball = �

λ = h/mu λ=

7.5A

1 kg

1 kg

� = 0.0397 kg

1000 g 55 mi 1 hr hr

��

��

1.609 km

1000 m

mi

1 km

3600 s

6.626x10–34 J•s h = = 6.67607x10–34 = 6.7x10–34 m (0.0397 kg)(25 m/s) mu

Plan: Use the equation ∆x • m∆u ≥

��

� = 24.5819 = 25 m/s

h to solve for the uncertainty (Δx) in position of the baseball.

4π

Solution:

h 4π Δu = 1% of u = 0.0100(44.7 m/s) = 0.447 m/s h 6.626x10−34 J • s ∆x ≥ = = 8.3070285x10– 34 4πm∆u 4π ( 0.142 kg )( 0.447 m/s ) Δx ≥ 8.31x10– 34 m ∆x • m∆u ≥

7.5B

Plan: Use the equation ∆x • m∆u ≥

h to solve for the uncertainty (Δx) in position of the neutron.

4π

Solution:

h 4π Δu = 1% of u = 0.0100(8x107 m/s) = 8x105 m/s ∆x • m∆u ≥

6.626x10–34 J•s h = = 3.9467x10– 14 m 4πm∆u 4π�1.67x10–27 kg�(8x105 m/s) Δx ≥ 4x10– 14 m

∆x ≥

7.6A

Plan: Following the rules for l (integer from 0 to n – 1) and m l (integer from –l to +l), write quantum numbers for n = 4.

Solution: For n = 4 l = 0, 1, 2, 3 For l = 0, m l = 0

7-3

For l = 1, m l = –1, 0, 1 For l = 2, m l = –2, –1, 0, 1, 2 For l = 3, m l = –3, –2, –1, 0, 1, 2, 3 7.6B

Plan: Following the rules for l (integer from 0 to n – 1) and m l (integer from –l to +l), determine which value of the principal quantum number has five allowed levels of l.

Solution: The number of possible l values is equal to n, so the n = 5 principal quantum number has five allowed values of l.

For n = 5 l = 0, 1, 2, 3, 4 For l = 0, m l = 0 For l = 1, m l = –1, 0, 1 For l = 2, m l = –2, –1, 0, 1, 2 For l = 3, m l = –3, –2, –1, 0, 1, 2, 3 For l = 4, m l = –4, –3, –2, –1, 0, 1, 2, 3, 4

7.7A

Plan: Identify n and l from the sublevel designation. n is the integer in front of the sublevel letter. The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, f represents l = 3.

Knowing the value for l, find the m l values (integer from –l to +l).

Solution: Sublevel name n value l value m l values 2p 2 1 –1, 0, 1 5f 5 3 –3, –2, –1, 0, 1, 2, 3

7.7B

Plan: Identify n and l from the sublevel designation. n is the integer in front of the sublevel letter. The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, f represents l = 3.

Knowing the value for l, find the m l values (integer from –l to +l).

Solution: Sublevel name n value l value m l values 4d 4 2 –2, –1, 0, 1, 2 6s 6 0 0 The number of orbitals for each sublevel equals 2l + 1. Sublevel 4d should have 5 orbitals and sublevel 6s should have 1 orbital. Both of these agree with the number of m l values for the sublevel.

7.8A

Plan: Use the rules for designating quantum numbers to fill in the blanks.

For a given n, l can be any integer from 0 to n–1.

For a given l, m l can be any integer from – l to + l.

The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, f represents l = 3.

Solution: The completed table is: n l ml Name a) 4 1 0 4p b) 2 1 0 2p c) 3 2 –2 3d d) 2 0 0 2s

7.8B

Plan: Use the rules for designating quantum numbers to determine what is wrong with the quantum number designations provided in the problem.

For a given n, l can be any integer from 0 to n–1.

For a given l, m l can be any integer from – l to + l.

The sublevels are given a letter designation, in which s represents l = 0, p represents l = 1, d represents l = 2, f represents l = 3.

7-4

Solution: The provided table is: n l ml Name a) 5 3 4 5f b) 2 2 1 2d c) 6 1 –1 6s a) For l = 3, the allowed values for m l are –3, –2, –1, 0, 1, 2, 3, not 4 b) For n = 2, l = 0 or 1 only, not 2; the sublevel is 2p, since m l = 1. c) The value l = 1 indicates the p sublevel, not the s; the sublevel name is 6p.

TOOLS OF THE LABORATORY BOXED READING PROBLEMS Plan: Plot absorbance on the y-axis and concentration on the x-axis. Since this is a linear plot, the graph is of the type y = mx + b, with m = slope and b = intercept. Any two points may be used to find the slope, and the slope is used to find the intercept. Once the equation for the line is known, the absorbance of the solution in part b) is used to find the concentration of the diluted solution, after which the dilution equation is used to find the molarity of the original solution.

Solution: a) Absorbance vs. Concentration:

Absorbance

B7.1

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0

0.00001

0.00002

0.00003

0.00004

Concentration (M)

This is a linear plot, thus, using the first and last points given: ( 0.396 − 0.131) y − y1 = = 13,250 = 1.3x104/M m= 2 5 5 − − x2 − x1 3.0x10 − 1.0x10 M

(

)

Using the slope just calculated and any of the data points, the value of the intercept may be found. b = y – mx = 0.396 – (13,250/M)(3.0x10–5 M) = –0.0015 = 0.00 (absorbance has no units) b) Use the equation just determined: y = (1.3x104/M) x + 0.00. x = (y – 0.00)/(1.3x104/M) = (0.236/1.3x104) M = 1.81538x10–5 M = 1.8x10–5 M This value is M f in a dilution problem (M i V i )= (M f V f ) with V i = 20.0 mL and V f = 150. mL. Mi = B7.2

−5 (M f )(Vf ) (1.81538x10 M ) (150. mL ) = = 1.361538x10–4 = 1.4x10–4 M ( 20.0 mL ) (Vi )

Plan: The color of light associated with each wavelength can be found from Figure 7.3. The c frequency of each wavelength can be determined from the relationship c = λν or ν = . The

λ

7-5

wavelength in nm must be converted to meters.

Solution: 3.00x108 m/s  1 nm  14 14 –1 a) red ν=   = 4.4709x10 = 4.47x10 s 671 nm  10−9 m  3.00x108 m/s  1 nm  14 14 –1 ν= b) blue  −9  = 6.6225x10 = 6.62x10 s 453 nm  10 m  3.00x108 m/s  1 nm  14 14 –1 c) yellow-orange ν =   = 5.0933786x10 = 5.09x10 s 589 nm  10−9 m  END–OF–CHAPTER PROBLEMS 7.1

All types of electromagnetic radiation travel as waves at the same speed. They differ in both their frequency, wavelength, and energy.

7.2

Plan: Recall that the shorter the wavelength, the higher the frequency and the greater the energy. Figure 7.3 describes the electromagnetic spectrum by wavelength and frequency.

Solution: a) Wavelength increases from left (10–2 nm) to right (1012 nm) in Figure 7.3. The trend in increasing wavelength is: x-ray < ultraviolet < visible < infrared < microwave < radio wave. b) Frequency is inversely proportional to wavelength according to the equation c = λν, so frequency has the opposite trend: radio wave < microwave < infrared < visible < ultraviolet < x-ray. c) Energy is directly proportional to frequency according to the equation E = hν. Therefore, the trend in increasing energy matches the trend in increasing frequency: radio wave < microwave < infrared < visible < ultraviolet < x-ray.

7.3

a) Refraction is the bending of light waves at the boundary of two media, as when light travels from air into water. b) Diffraction is the bending of light waves around an object, as when a wave passes through a slit about as wide as its wavelength. c) Dispersion is the separation of light into its component colors (wavelengths), as when light passes through a prism. d) Interference is the bending of light through a series of parallel slits to produce a diffraction pattern of brighter and darker spots.

Note: Refraction leads to a dispersion effect and diffraction leads to an interference effect.

7.4

Evidence for the wave model is seen in the phenomena of diffraction and refraction. Evidence for the particle model includes the photoelectric effect and blackbody radiation.

7.5

a) Frequency: C < B < A b) Energy: C < B < A c) Amplitude: B < C < A d) Since wave A has a higher energy and frequency than B, wave A is more likely to cause a current. e) Wave C is more likely to be infrared radiation since wave C has a longer wavelength than B.

7.6

Radiation (light energy) occurs as quanta of electromagnetic radiation, where each packet of energy is called a photon. The energy associated with this photon is fixed by its frequency, E = hν. Since energy depends on frequency, a threshold (minimum) frequency is to be expected. A current will flow as soon as a photon of sufficient energy reaches the metal plate, so there is no time lag.

7.7

Plan: Wavelength is related to frequency through the equation c = λν. Recall that a Hz is a reciprocal second, or 1/s = s–1. Assume that the number “950” has three significant figures.

7-6

Solution: c = λν

λ (m) =

c = ν

λ (nm) = λ (Å) =

7.8

3.00x108 m/s = 315.789 = 316 m  103 Hz   s −1  ( 950. kHz )      1 kHz   Hz 

c  1 nm  11 11 = ( 315.789 m )  −9  = 3.15789x10 = 3.16x10 nm ν  10 m 

 1Å  c = ( 315.789 m )  −10  = 3.158x1012 = 3.16x1012 Å ν  10 m 

Wavelength and frequency relate through the equation c = λν. Recall that a Hz is a reciprocal second, or 1/s = s–1.

λ (m) =

c = ν

λ (nm) = λ (Å) =

3.00x108 m/s = 3.208556 = 3.21 m  106 Hz   s −1  ( 93.5 MHz )      1 MHz   Hz 

c  1 nm  9 9 = ( 3.208556 m )  −9  = 3.208556x10 = 3.21x10 nm ν  10 m 

 1Å  c = ( 3.208556 m )  −10  = 3.208556x1010 = 3.21x1010 Å ν  10 m 

7.9

Plan: Frequency is related to energy through the equation E = hν. Note that 1 Hz = 1 s–1.

Solution: E = hν E = (6.626x10–34 J•s)(3.8x1010 s–1) = 2.51788x10–23 = 2.5x10–23 J

7.10

E=

7.11

Plan: Energy is inversely proportional to wavelength ( E =

hc

=

λ

(6.626x10

−34

)(

)

J•s 3.00x108 m/s  1 Å  –15 –15  −10  = 1.5291x10 = 1.5x10 J 1.3 Å 10 m  

hc

λ

). As wavelength decreases, energy increases.

Solution: In terms of increasing energy the order is red < yellow < blue.

7.12

Since energy is directly proportional to frequency (E = hν): UV (ν = 8.0x1015 s–1) > IR (ν = 6.5x1013 s–1) > microwave (ν = 9.8x1011 s–1) or UV > IR > microwave.

7.13

Plan: Wavelength is related to frequency through the equation c = λν. Recall that a Hz is a reciprocal second, or 1/s = s–1.

Solution:  109 Hz   s −1  ν = (s–1) = ( 22.235 GHz )  = 2.2235x1010 s–1  1 GHz   Hz    

λ (nm) = λ (Å) =

2.9979x108 m/s  1 nm  c 7 7 =   = 1.3482797x10 = 1.3483x10 nm ν 2.2235x1010 s −1  10−9 m 

c 2.9979x108 m/s  1 Å  8 8 =   = 1.3482797x10 = 1.3483x10 Å ν 2.2235x1010 s −1  10−10 m 

7-7

7.14

Frequency and wavelength can be calculated using the speed of light: c = λν.

3.00x108 m/s  1 µm  c 13 13 –1 = a) ν =  −6  = 3.125x10 = 3.1x10 s 9.6 µm λ  10 m   1 µm  c 2.9979x108 m/s b) λ (µm) = =   = 3.464979 = 3.465 µm − 1 ν  s   10−6 m  13 8.652x10 Hz    Hz 

(

7.15

)

Frequency and energy are related by E = hν, and wavelength and energy are related by E = hc/λ.  106 eV   1.602x10−19 J  (1.33 MeV )     Hz 1 eV E  20 20  1 MeV    ν (Hz) = =  −1  = 3.2156x10 = 3.22x10 Hz h 6.626x10−34 J•s s 

λ (m) =

hc = E

( 6.626x10

−34

)(

J•s 3.00x108 m/s

)

= 9.32950x10–13 = 9.33x10–13 m  10 eV   1.602x10 J  (1.33 MeV )     1 eV  1 MeV    The wavelength can also be found using the frequency calculated in the equation c = λν.

7.16

−19

6

Plan: The least energetic photon in part a) has the longest wavelength (242 nm). The most energetic photon in part b) has the shortest wavelength (2200 Å). Use the relationship c = λν to find the frequency of the photons and hc relationship E = to find the energy.

λ

Solution: a) c = λν

ν=

c

λ

=

3.00x108 m/s  1 nm  15 15 –1   = 1.239669x10 = 1.24x10 s 242 nm  10−9 m 

( 6.626x10

−34

)(

)

−34

)(

)

J•s 3.00x108 m/s  1 nm  –19 –19  −9  = 8.2140x10 = 8.21x10 J 242 nm λ  10 m  8 3.00x10 m/s  1 Å  c 15 15 –1 = b) ν =   = 1.3636x10 = 1.4x10 s 2200 Å  10−10 m  λ

E=

E=

hc

hc

λ

=

=

(6.626x10

J•s 3.00x108 m/s  1 Å  –19 –19  −10  = 9.03545x10 = 9.0x10 J 2200 Å 10 m  

7.17

“n” in the Rydberg equation is equal to a Bohr orbit of quantum number “n” where n = 1, 2, 3, ...∞.

7.18

Bohr’s key assumption was that the electron in an atom does not radiate energy while in a stationary state, and the electron can move to a different orbit by absorbing or emitting a photon whose energy is equal to the difference in energy between two states. These differences in energy correspond to the wavelengths in the known spectra for the hydrogen atoms. A Solar System model does not allow for the movement of electrons between levels.

7.19

An absorption spectrum is produced when atoms absorb certain wavelengths of incoming light as electrons move from lower to higher energy levels and results in dark lines against a bright background. An emission spectrum is produced when atoms that have been excited to higher energy emit photons as their electrons return to lower energy levels and results in colored lines against a dark background. Bohr worked with emission spectra.

7.20

Plan: The quantum number n is related to the energy level of the electron. An electron absorbs energy to change from lower energy (lower n) to higher energy (higher n), giving an absorption spectrum. An electron emits energy as it drops from a higher energy level (higher n) to a lower one (lower n), giving an emission spectrum.

7-8

Solution: a) The electron is moving from a lower value of n (2) to a higher value of n (4): absorption b) The electron is moving from a higher value of n (3) to a lower value of n (1): emission c) The electron is moving from a higher value of n (5) to a lower value of n (2):emission d) The electron is moving from a lower value of n (3) to a higher value of n (4): absorption 7.21

The Bohr model works only for a one-electron system. The additional attractions and repulsions in many-electron systems make it impossible to predict accurately the spectral lines.

7.22

The Bohr model has successfully predicted the line spectra for the H atom and Be3+ ion since both are one-

( )( 2.18x10 J ) where Z is the atomic number

− Z2 electron species. The energies could be predicted from E n =

−18

n2 for the atom or ion. The line spectra for H would not match the line spectra for Be3+ since the H nucleus contains one proton while the Be3+ nucleus contains 4 protons (the Z values in the equation do not match); the force of attraction of the nucleus for the electron would be greater in the beryllium ion than in the hydrogen atom. This means that the pattern of lines would be similar, but at different wavelengths.

7.23

Plan: Calculate wavelength by substituting the given values into Equation 7.3, where n 1 = 2 and n 2 = 5 because n 2 > n 1 . Although more significant figures could be used, five significant figures are adequate for this calculation.

Solution:  1 1 1  R = 1.096776x107 m–1 = R 2 − 2    λ n n 2   1 n1 = 2 n2 = 5  1 1 1  1   1 = R  2 − 2  = 1.096776x107 m −1  2 − 2  = 2,303,229.6 m–1 n  λ 2 5 n   2   1

(

)

   1 nm  1 λ (nm) =   = 434.1729544 = 434.17 nm −1   −9  2,303, 229.6 m   10 m  7.24

Calculate wavelength by substituting the given values into the Rydberg equation, where n 1 = 1 and n 2 = 3 because n 2 > n 1 . Although more significant figures could be used, five significant figures are adequate for this calculation.  1 1 1  1  1 = R  2 − 2  = 1.096776x107 m −1  2 − 2  = 9,749,120 m–1   λ 3  n2  1  n1

(

)

   1Å  1 λ (Å) =   = 1025.7336 = 1025.7 Å −1   −10  9,749,120 m   10 m  Plan: The Rydberg equation is needed. For the infrared series of the H atom, n 1 equals 3. The least energetic spectral line in this series would represent an electron moving from the next highest energy level, n 2 = 4.

Although more significant figures could be used, five significant figures are adequate for this calculation.

Solution:  1 1 1  1   1 = R  2 − 2  = 1.096776x107 m −1  2 − 2  = 533,155 m–1 n  λ 4  n2  3  1 7.25

(

)

   1 nm  1 λ (nm) =   = 1875.627 = 1875.6 nm −1   −9  533,155 m   10 m  7.26

Plan: The Rydberg equation is needed. For the visible series of the H atom, n 1 equals 2. The least energetic spectral line in this series would represent an electron moving from the next highest energy level, n = 3. Although more significant figures could be used, five significant figures are adequate for this calculation.

7-9

Solution:  1 1 1  1   1 = R  2 − 2  = 1.096776x107 m −1  2 − 2  = 1,523,300 m–1 n  λ 2 3 n   2   1

(

)

   1 nm  1 λ (nm) =   = 656.4695 = 656.47 nm −1   −9  1,523,300 m   10 m  7.27

7.28

7.29

Plan: To find the transition energy, use the equation for the energy of an electron transition and multiply by Avogadro’s number to convert to energy per mole.

Solution:  1 1  ∆E = −2.18x10−18 J  2 − 2  n ninitial   final 1   1 ∆E = −2.18x10−18 J  2 − 2  = –4.578x10–19 J/photon 5  2 − 19  −4.578x10 J   6.022x1023 photons  5 5 ∆E =     = –2.75687x10 = –2.76x10 J/mol   photon 1 mol    The energy has a negative value since this electron transition to a lower n value is an emission of energy.

(

)

(

)

To find the transition energy, use the equation for the energy of an electron transition and multiply by Avogadro’s number.  1 1  ∆E = −2.18x10−18 J  2 − 2  n ninitial   final 1  1 ∆E = −2.18x10−18 J  2 − 2  = 1.93778x10–18 J/photon 1  3 − 18  1.93778x10 J   6.022x1023 photons  6 6 ∆E =     = 1.1669x10 = 1.17x10 J/mol  photon 1 mol   

(

)

(

)

Plan: Determine the relative energy of the electron transitions. Remember that energy is directly proportional to frequency (E = hν).

Solution: Looking at an energy chart will help answer this question. n=5 n=4 (d) (a)

n=3

(c) n=2 (b) n=1

7.30

Frequency is proportional to energy so the smallest frequency will be d) n = 4 to n = 3; levels 3 and 4 have a smaller ∆E than the levels in the other transitions. The largest frequency is b) n = 2 to n = 1 since levels 1 and 2 have a larger ∆E than the levels in the other transitions. Transition a) n = 2 to n = 4 will be smaller than transition c) n = 2 to n = 5 since level 5 is a higher energy than level 4. In order of increasing frequency the transitions are d < a < c < b. b>c>a>d

7-10

7.31

Plan: Use the Rydberg equation. Since the electron is in the ground state (lowest energy level), n 1 = 1. Convert the wavelength from nm to units of meters.

Solution:  10−9 m  λ = ( 97.20 nm )  = 9.720x10–8 m ground state: n 1 = 1; n 2 = ?  1 nm     1 1 1  = 1.096776x107 m −1  2 − 2  n λ n2   1

)

(

1 1 1  = 1.096776x107 m −1  2 − 2  −8  9.72x10 m n2  1 1 1  0.93803 =  2 − 2  1 n2   1 = 1 – 0.93803 = 0.06197 n22

(

)

n22 = 16.14 n2 = 4

7.32

 10−9 m  = 1.281x10–6 m  1 nm   

λ = (1281 nm ) 

 1 1  = 1.096776x107 m −1  2 − 2   λ n2   n1

1

(

)

 1 1 1  = 1.096776x107 m −1  2 − 2  −6  1.281x10 m 5   n1

(

)

 1 1  0.07118 =  2 − 2  n 5   1 1 = 0.07118 + 0.04000 = 0.11118 n12 n12 = 8.9944 n1 = 3

hc

( 6.626x10 =

−34

)(

)

J•s 3.00x108 m/s  1 nm  –19 –19  −9  = 4.55917x10 = 4.56x10 J ( 436 nm ) 10 m  

7.33

E=

7.34

a) Absorptions: A, C, D; Emissions: B, E, F b) Energy of emissions: E < F < B c) Wavelength of absorption: D < A < C

7.35

If an electron occupies a circular orbit, only integral numbers of wavelengths (= 2nπr) are allowed for acceptable standing waves. A wave with a fractional number of wavelengths is forbidden due to destructive interference with itself. In a musical analogy to electron waves, the only acceptable guitar string wavelengths are those that are an integral multiple of twice the guitar string length (2 L).

7.36

De Broglie’s concept is supported by the diffraction properties of electrons demonstrated in an electron microscope.

λ

7-11

7.37

Macroscopic objects have significant mass. A large m in the denominator of λ = h/mu will result in a very small wavelength. Macroscopic objects do exhibit a wavelike motion, but the wavelength is too small for humans to see it.

7.38

The Heisenberg uncertainty principle states that there is fundamental limit to the accuracy of measurements. This limit is not dependent on the precision of the measuring instruments, but is inherent in nature.

7.39

Plan: Use the de Broglie equation. Mass in lb must be converted to kg and velocity in mi/h must be converted to m/s because a joule is equivalent to kg•m2/s2.

Solution:  1 kg  a) Mass (kg) = ( 232 lb )   = 105.2154 kg  2.205 lb  3  19.8 mi   1 km   10 m   1 h  Velocity (m/s) =       = 8.87097 m/s   h   0.62 mi   1 km   3600 s 

λ=

h = mu

( 6.626x10 J•s ) −34

 kg•m 2 /s2  –37 –37   = 7.099063x10 = 7.10x10 m m   J   (105.2154 kg )  8.87097  s  

3  0.1 mi   1 km   10 m   1 h  b) Uncertainty in velocity (m/s) =       = 0.0448029 m/s   h   0.62 mi   1 km   3600 s 

∆x•m∆v ≥ ∆x ≥

7.40

h 4π

h ≥ 4 π m∆v

a) λ =

h = mu

( 6.626x10 J•s ) −34

 kg•m 2 /s2  –35 –35   ≥ 1.11855x10 ≥ 1x10 m J  0.0448029 m    4π (105.2154 kg )   s  

( 6.626x10 J•s )  kg•m /s   10 g   0.62 mi   1 km   3600 s        ( 6.6x10 g )  3.4x10 mih   J   1 kg   1 km   10 m   1 h  −34

2

−24

2

3

3

7

= 6.59057x10–15 = 6.6x10–15 m b) ∆x•m∆v ≥

h 4π

h ≥ ∆x ≥ 4 π m∆v

 kg•m /s   10 g   0.62 mi   1 km   3600 s  (6.626x10 J•s )     1 kg   1 km   10 m   1 h  J  0.1x10 mi        4π ( 6.6x10 g )   h −34

−24

≥ 1.783166x10

–14

7.41

2

2

3

3

7

  –14 ≥ 2x10 m

Plan: Use the de Broglie equation. Mass in g must be converted to kg and wavelength in Ǻ must be converted to m because a joule is equivalent to kg•m2/s2.

Solution:  1 kg  Mass (kg) = ( 56.5 g )  3  = 0.0565 kg  10 g     10−10 m  = 5.4x10–7 m Wavelength (m) = ( 5400 Å )   1 Å   

7-12

7.42

7.43

7.44

λ=

h mu

u=

6.626x10−34 J•s  kg•m 2 /s2  h –26 –26 =   = 2.1717x10 = 2.2x10 m/s 7 −   mλ J ( 0.0565 kg ) 5.4x10 m  

λ=

h mu

u=

6.626x10 J•s  kg•m 2 /s2   103 g   1 pm  h = = 4.666197x10–23 = 4.67x10–23 m/s    mλ (142 g )(100. pm )  J   1 kg   10−12 m 

(

)

(

(

)

)

−34

Plan: The de Broglie wavelength equation will give the mass equivalent of a photon with known wavelength and velocity. The term “mass equivalent” is used instead of “mass of photon” because photons are quanta of electromagnetic energy that have no mass. A light photon’s velocity is the speed of light, 3.00x108 m/s.

Wavelength in nm must be converted to m.

Solution:  10−9 m  Wavelength (m) = ( 589 nm )  = 5.89x10–7 m  1 nm    h λ= mu

(

)

−34

m=

6.626x10 J•s  kg•m 2 /s2  h –36 –36 =   = 3.7499x10 = 3.75x10 kg/photon 8 −7  λu J 5.89x10 m 3.00 x10 m/s  

λ=

h mu

m=

6.626x10 J•s  kg•m 2 /s2   1 nm  h –36 =    −9  = 3.2916x10 kg/photon 8 λu J 10 m ( 671 nm ) 3.00x10 m/s   

(

(

)(

−34

(

)

)

)

 3.2916x10−36 kg   6.022x1023 photons  –12 –12     = 1.9822x10 = 1.98x10 kg/mol photon mol   

7.45

The quantity ψ2 expresses the probability of finding an electron within a specified tiny region of space.

7.46

Since ψ2 is the probability of finding an electron within a small region or volume, electron density would represent a probability per unit volume and would more accurately be called electron probability density.

7.47

A peak in the radial probability distribution at a certain distance means that the total probability of finding the electron is greatest within a thin spherical volume having a radius very close to that distance. Since principal quantum number (n) correlates with distance from the nucleus, the peak for n = 2 would occur at a greater distance from the nucleus than 0.529 Å. Thus, the probability of finding an electron at 0.529 Å is much greater for the 1s orbital than for the 2s.

7.48

a) Principal quantum number, n, relates to the size of the orbital. More specifically, it relates to the distance from the nucleus at which the probability of finding an electron is greatest. This distance is determined by the energy of the electron. b) Angular momentum quantum number, l, relates to the shape of the orbital. It is also called the azimuthal quantum number.

7-13

c) Magnetic quantum number, m l , relates to the orientation of the orbital in space in three-dimensional space.

7.49

Plan: The following letter designations correlate with the following l quantum numbers: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that allowed m l values are – l to + l.

The number of orbitals of a particular type is given by the number of possible m l values.

Solution: a) There is only a single s orbital in any shell. l = 1 and m l = 0: one value of m l = one s orbital. b) There are five d orbitals in any shell. l = 2 and m l = –2, –1, 0, +1, +2. Five values of m l = five d orbitals. c) There are three p orbitals in any shell. l = 1 and m l = –1, 0, +1. Three values of m l = three p orbitals. d) If n = 3, l = 0(s), 1(p), and 2(d). There is a 3s (1 orbital), a 3p set (3 orbitals), and a 3d set (5 orbitals) for a total of nine orbitals (1 + 3 + 5 = 9).

7.50

a) All f orbitals consist of sets of seven (l = 3 and m l = –3, –2, –1, 0, +1, +2, +3). b) All p orbitals consist of sets of three (l = 1 and m l = –1, 0, +1). c) All d orbitals consist of sets of five (l = 2 and m l = –2, –1, 0, +1, +2). d) If n = 2, then there is a 2s (1 orbital) and a 2p set (3 orbitals) for a total of four orbitals (1 + 3 = 4).

7.51

Plan: Magnetic quantum numbers (m l ) can have integer values from –l to + l. The l quantum number can have integer values from 0 to n – 1.

Solution: a) l = 2 so m l = –2, –1, 0, +1, +2 b) n = 1 so l = 1 – 1 = 0 and m l = 0 c) l = 3 so m l = –3, –2, –1, 0, +1, +2, +3

7.52

Magnetic quantum numbers can have integer values from –l to +l. The l quantum number can have integer values from 0 to n – 1. a) l = 3 so m l = –3, –2, –1, 0, +1, +2, +3 b) n = 2 so l = 0 or 1; for l = 0, m l = 0; for l = l, m l = –1,0,+1 c) l = 1 so m l = –1, 0, +1

7.53

Plan: The s orbital is spherical; p orbitals have two lobes; the subscript x indicates that this orbital lies along the x-axis.

Solution: a) s: spherical b) p x : 2 lobes along the x-axis

z

z

x

x

y

y

The variations in coloring of the p orbital are a consequence of the quantum mechanical derivation of atomic orbitals that are beyond the scope of this course.

7.54

a) p z : 2 lobes along the z-axis

b) d xy : 4 lobes

7-14

z

y

y

x

x

The variations in coloring of the p and d orbitals are a consequence of the quantum mechanical derivation of atomic orbitals that are beyond the scope of this course.

7.55

Plan: The following letter designations for the various sublevels (orbitals) correlate with the following l quantum numbers: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that allowed m l values are – l to + l. The number of orbitals of a particular type is given by the number of possible m l values.

Solution: sublevel allowable m l # of possible orbitals a) d (l = 2) –2, –1, 0, +1, +2 5 b) p (l = 1) –1, 0, +1 3 c) f (l = 3) –3, –2, –1, 0, +1, +2, +3 7

7.56

sublevel a) s (l = 0) b) d (l = 2) c) p (l = 1)

7.57

Plan: The integer in front of the letter represents the n value. The letter designates the l value: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that allowed m l values are – l to + l.

Solution: a) For the 5s subshell, n = 5 and l = 0. Since m l = 0, there is one orbital. b) For the 3p subshell, n = 3 and l = 1. Since m l = –1, 0, +1, there are three orbitals. c) For the 4f subshell, n = 4 and l = 3. Since m l = –3, –2, –1, 0, +1, +2, +3, there are seven orbitals.

7.58

a) n = 6; l = 4; 9 orbitals (m l = –4, –3, –2, –1, 0, +1, +2, +3, +4) b) n = 4; l = 0; 1 orbital (m l = 0) c) n = 3; l = 2; 5 orbitals (m l = –2, –1, 0, +1, +2)

7.59

Plan: Allowed values of quantum numbers: n = positive integers; l = integers from 0 to n – 1; m l = integers from – l through 0 to + l.

Solution: a) n = 2; l = 0; m l = –1: With n = 2, l can be 0 or 1; with l = 0, the only allowable m l value is 0. This combination is not allowed. To correct, either change the l or m l value.

Correct: n = 2; l = 1; m l = –1 or n = 2; l = 0; m l = 0. b) n = 4; l = 3; m l = –1: With n = 4, l can be 0, 1, 2, or 3; with l = 3, the allowable m l values are –3, –2, –1, 0, +1, +2, +3. Combination is allowed. c) n = 3; l = 1; m l = 0: With n = 3, l can be 0, 1, or 2; with l = 1, the allowable m l values are –1, 0, +1.

Combination is allowed. d) n = 5; l = 2; m l = +3: With n = 5, l can be 0, 1, 2, 3, or 4; with l = 2, the allowable m l values are –2, –1, 0, +1, +2. +3 is not an allowable m l value. To correct, either change l or m l value.

Correct: n = 5; l = 3; m l = +3 or n = 5; l = 2; m l = 0.

allowable m l 0 –2, –1, 0, +1, +2 –1, 0, +1

# of possible orbitals 1 5 3

7-15

7.60

a) Combination is allowed. b) No; n = 2; l = 1; m l = +1 or n = 2; l = 1; m l = 0 c) No; n = 7; l = 1; m l = +1 or n = 7; l = 3; m l = 0 d) No; n = 3; l = 1; m l = –1 or n = 3; l = 2; m l = –2

7.61

Determine the λ max for β-carotene by measuring its absorbance in the 610-640 nm region of the visible spectrum.

Prepare a series of solutions of β-carotene of accurately known concentration (using benzene or chloroform as solvent), and measure the absorbance for each solution. Prepare a graph of absorbance versus concentration for these solutions and determine its slope (assuming that this material obeys Beer’s law). Measure the absorbance of the oil expressed from orange peel (diluting with solvent if necessary). The β-carotene concentration can then either be read directly from the calibration curve or calculated from the slope (A = kC, where k = slope of the line and C = concentration).

7.62

Plan: For Ppart a, use the values of the constants h, π, m e , and a 0 to find the overall constant in the equation. hc Use the resulting equation to calculate ∆E in part b). Use the relationship E = to calculate the wavelength in

λ

part c). Remember that a joule is equivalent to kg•m2/s2.

Solution: a) h = 6.626x10–34 J•s; m e = 9.1094x10–31 kg; a 0 = 52.92x10–12 m E= −

E= −

h2  1  − = 2 2 2 2 2  2  8π me a0 n 8π me a0  n  h2

( 6.626x10

(

−34

J•s

)(

)

2

8π 2 9.1094x10−31 kg 52.92x10−12 m

)

 kg•m 2 /s2   1    2  J   n 

2  

 1   1  = –(2.17963x10–18 J)  2  = –(2.180x10–18 J)  2  n  n  This is identical with the result from Bohr’s theory. For the H atom, Z = 1 and Bohr’s constant = –2.18x10–18 J. For the hydrogen atom, derivation using classical principles or quantum-mechanical principles yields the same constant. b) The n = 3 energy level is higher in energy than the n = 2 level. Because the zero point of the atom’s energy is defined as an electron’s infinite distance from the nucleus, a larger negative number describes a lower energy level. Although this may be confusing, it makes sense that an energy change would be a positive number.

1   1 ∆E = –(2.180x10–18 J)  2 − 2  = –3.027778x10–19 = 3.028x10–19 J 3  2 hc c) E =

λ

λ (m) =

(

)(

)

6.626x10−34 J•s 2.9979x108 m/s hc = = 6.56061x10–7 = 6.561x10–7 m −19 E 3.027778x10 J

(

(



)



) 1 nm  = 656.061 = 656.1 nm

λ (nm) = 6.56061x10−7 m 

−9

 10 m  This is the wavelength for the observed red line in the hydrogen spectrum.

7.63

Plan: When light of sufficient frequency (energy) shines on metal, electrons in the metal break free and a current flows.

Solution:

7-16

a) The lines do not begin at the origin because an electron must absorb a minimum amount of energy before it has enough energy to overcome the attraction of the nucleus and leave the atom. This minimum energy is the energy of photons of light at the threshold frequency. b) The lines for K and Ag do not begin at the same point. The amount of energy that an electron must absorb to leave the K atom is less than the amount of energy that an electron must absorb to leave the Ag atom, where the attraction between the nucleus and outer electron is stronger than in a K atom. c) Wavelength is inversely proportional to energy. Thus, the metal that requires a larger amount of energy to be absorbed before electrons are emitted will require a shorter wavelength of light. Electrons in Ag atoms require more energy to leave, so Ag requires a shorter wavelength of light than K to eject an electron. d) The slopes of the line show an increase in kinetic energy as the frequency (or energy) of light is increased.

Since the slopes are the same, this means that for an increase of one unit of frequency (or energy) of light, the increase in kinetic energy of an electron ejected from K is the same as the increase in the kinetic energy of an electron ejected from Ag. After an electron is ejected, the energy that it absorbs above the threshold energy becomes the kinetic energy of the electron. For the same increase in energy above the threshold energy, for either K or Ag, the kinetic energy of the ejected electron will be the same.

7.64

( 6.626x10

)

−34

J•s)(3.00x108 m/s  1 nm  –19  −9  = 2.8397x10 J 700. nm λ  10 m  This is the value for each photon, that is, J/photon.

1 photon   = 70.430 = 70. photons Number of photons = 2.0x10−17 J  −19  J  2.8397x10

a) E =

hc

=

(

hc

)

(6.626x10 J•s)(3.00x10 m/s )  1nm  = 4.18484x10 J = −34

8

–19

 −9  475. nm  10 m  This is the value for each photon, that is, J/photon.

b) E =

λ

(

)

1 photon   = 47.7916 = 48 photons Number of photons = 2.0x10−17 J  −19  4.18484x10 J  

7.65

Determine the wavelength: λ = 1/(1953 cm–1) = 5.1203277x10–4 cm  10−2 m   1 nm  λ (nm) = 5.1203277x10−4 cm  = 5120.3277 = 5.120x103 nm  1 cm   10−9 m   

(

)



(

ν = c/λ = 7.66

−2





) 101 cmm   101 Åm  = 51203.277 = 5.120x10 Å

λ (Å) = 5.1203277x10−4 cm   



−10

4



2.9979x10 m/s  1 cm   1 Hz  13 13    = 5.8548987x10 = 5.855x10 Hz 5.1203277x10−4 cm  10−2 m   1 s −1  8

Plan: The Bohr model has been successfully applied to predict the spectral lines for one-electron species other than H. Common one-electron species are small cations with all but one electron removed. Since the problem specifies a metal ion, assume that the possible choices are Li2+ or Be3+. Use the relationship E = hν to convert the  Z2  frequency to energy and then solve Bohr’s equation E = 2.18x10−18 J  2  to verify if a whole number for Z n    can be calculated. Recall that the negative sign is a convention based on the zero point of the atom’s energy; it is deleted in this calculation to avoid taking the square root of a negative number.

Solution: The highest energy line corresponds to the transition from n = 1 to n = ∞.

E = hν = (6.626x10–34 J•s) (2.961x1016 Hz) (s–1/Hz) = 1.9619586x10–17 J

(

7-17

)

 Z2  E = 2.18x10−18 J  2  n   

(

Z2 =

)

Z = charge of the nucleus

1.9619586x10−17 (12 ) En 2 = = 8.99998 2.18x10−18 J 2.18x10 −18 J

Then Z2 = 9 and Z = 3.

Therefore, the ion is Li2+ with an atomic number of 3.

7.67

( 6.626 x10 J•s )  kg•m /s  = 2.139214x10 = 2.1x10 m   ( 9.11x10 kg )  3.4x10 ms   J  ( 6.626x10 J•s )  kg•m /s  = 1.16696x10 = 1.2x10 m h = Proton: λ =   mu (1.67x10 kg )  3.4x10 ms   J 

a) Electron: λ =

−34

h = mu

2

2

–10

−31

6

−34

2

2

–13

−27

b) E = 1/2mu2 u=

–10

–13

6

therefore u2 = 2E/m

2E m

Electron: u =

(

) )

2 2.7x10−15 J  kg•m 2 /s2  7   = 7.6991x10 m/s −31 J 9.11x10 kg  

( 2 ( 2.7x10 J )  kg•m /s  Proton: u =   = 1.7982x10 m/s (1.67x10 kg )  J   kg•m /s  ( 6.626x10 J•s ) h Electron: λ = =   = 9.44698x10 = 9.4x10 m m   J mu   9.11x10 kg 7.6991x10 ( )  s   kg•m /s  ( 6.626x10 J•s ) h = Proton: λ =   = 2.20646 x 10 = 2.2 x 10 m m   J mu   (1.67x10 kg ) 1.7982x10 s  −15

2

2

6

−27

−34

2

−31

–12

–12

–13

–13

7

−34

−27

7.68

2

2

2

6

Plan: The electromagnetic spectrum shows that the visible region goes from 400 to 750 nm (4000 Å to 7500 Å).

Thus, wavelengths b, c, and d are for the three transitions in the visible series with n final = 2. Wavelength a is in the ultraviolet region of the spectrum and the ultraviolet series has n final = 1. Wavelength e is in the infrared region of the spectrum and the infrared series has n final = 3. Use the Rydberg equation to find the n initial for each line.

Convert the wavelengths from Å to units of m.

Solution: n = ? → n = 1; λ = 1212.7 Å (shortest λ corresponds to the largest ∆E)  10−10 m  λ (m) = (1212.7 ) = 1.2127x10–7 m  1     1 1 1  = 1.096776x107 m −1  2 − 2  n λ n2   1

(

)

1  1   7 −1  1   = 1.096776x10 m  2 − 2  −7 n2   1.2127x10 m  1

(

)

7-18

1 1  0.7518456 =  2 − 2  1 n2    1   2  = 1 – 0.7518456  n2   1   2  = 0.2481544  n2  n 2 = 2 for line (a) (n = 2 → n = 1) n = ? → n = 3; λ = 10,938 Å (longest λ corresponds to the smallest ∆E)  10−10 m  λ (m) = (10,938 ) = 1.0938x10–6 m  1   

 1 1  = 1.096776x107 m −1  2 − 2   λ n2   n1

1

(

)

1  1   7 −1  1   = 1.096776x10 m  2 − 2  −6 n2   1.0938x10 m  3

(

)

 1 1  0.083357397 =  2 − 2  3 n2  

0.083357397 = 0.111111111 −

1 n22

 1   2  = 0.11111111 – 0.083357396  n2   1   2  = 0.0277537151  n2  n22 =36.03121

n 2 = 6 for line (e) (n = 6 → n = 3) For the other three lines, n 1 = 2.

For line (d), n 2 = 3 (largest λ → smallest ∆E).

For line (b), n 2 = 5 (smallest λ → largest ∆E).

For line (c), n 2 = 4.

7.69

E=

hc

λ

a) λ (nm) =

thus λ =

hc E

)(

)

(

)(

)

(

)(

)

(

6.626x10−34 J•s 3.00x108 m/s  1 nm  hc =  −9  = 432.130 = 432 nm E 4.60x10−19 J  10 m 

6.626x10−34 J•s 3.00x108 m/s  1 nm  hc b) λ (nm) = =  −9  = 286.4265 = 286 nm E 6.94x10−19 J  10 m  c) λ (nm) =

7.70

6.626x10−34 J•s 3.00x108 m/s  1 nm  hc =  −9  = 450.748 = 451 nm E 4.41x10−19 J  10 m 

Index of refraction = c/v; v = c/(index of refraction) a) Water v = c/(index of refraction) = (3.00x108 m/s)/(1.33) = 2.2556x108 = 2.26x108 m/s b) Diamond v = c/(index of refraction) = (3.00x108 m/s)/(2.42) = 1.239669x108 = 1.24x108 m/s

7-19

7.71

Extra significant figures are necessary because of the data presented in the problem.

He–Ne λ = 632.8 nm ν = 6.148x1014 s–1 Ar Ar–Kr E = 3.499x10–19 J λ = 663.7 nm Dye Calculating missing λ values: λ = c/ν = (2.9979x108 m/s)/(6.148x1014 s–1) = 4.8762199x 10–7 = 4.876x10–7 m Ar Ar–Kr λ = hc/E = (6.626x10–34 J•s) (2.9979x108 m/s)/(3.499x10–19 J) = 5.67707x10–7 = 5.677x10–7 m Calculating missing ν values: He–Ne ν = c/λ = (2.9979x108 m/s)/[632.8 nm (10–9 m/nm)] = 4.7375 x 1014 = 4.738 x 1014 s–1 Ar–Kr ν = E/h = (3.499x10–19 J)/(6.626x10–34 J•s) = 5.28071x1014 = 5.281x1014 s–1 ν = c/λ = (2.9979x108 m/s)/[663.7 nm (10–9 m/nm)] = 4.51695x1014 = 4.517x1014 s–1 Dye Calculating missing E values: He–Ne E = hc/λ = [(6.626x10–34 J•s)(2.9979x108 m/s)]/[632.8 nm (10–9 m/nm)] = 3.13907797x10–19 = 3.139x10–19 J Ar E = hν = (6.626x10–34 J•s)(6.148x1014 s–1) = 4.0736648x10–19 = 4.074x10–19 J Dye E = hc/λ = [(6.626x10–34 J•s)(2.9979x108 m/s)]/[663.7 nm (10–9 m/nm)] = 2.99293x10–19 = 2.993x10–19 J The colors may be predicted from Figure 7.3 and the frequencies.

Orange He–Ne ν = 4.738x1014 s–1 ν = 6.148x1014 s–1 Green Ar Yellow Ar–Kr ν = 5.281x1014 s–1 ν = 4.517x1014 s–1 Red Dye

7.72

Plan: Allowed values of quantum numbers: n = positive integers; l = integers from 0 to n – 1; m l = integers from – l through 0 to + l.

Solution: a) The l value must be at least 1 for m l to be – 1, but cannot be greater than n – 1 = 3 –1 = 2. Increase the l value to 1 or 2 to create an allowable combination. b) The l value must be at least 1 for m l to be +1, but cannot be greater than n – 1 = 3 – 1 = 2. Decrease the l value to 1 or 2 to create an allowable combination. c) The l value must be at least 3 for m l to be +3, but cannot be greater than n – 1 = 7 – 1 = 6. Increase the l value to 3, 4, 5, or 6 to create an allowable combination. d) The l value must be at least 2 for m l to be –2, but cannot be greater than n – 1 = 4 – 1 = 3. Increase the l value to 2 or 3 to create an allowable combination.

7.73

 1 1  = 1.096776x107 m −1  2 − 2   λ n2   n1

1

a)

(

)

1 1  1 nm  1  = 1.096776x107 m −1  2 − 2    −9 1 94.91 nm  10 m  n2  

(

)

1 1  0.9606608 =  2 − 2  1 n2   n2 = 5  1 nm  1 1  7 −1  1 b)   = 1.096776x10 m  2 − 2  1281 nm  10−9 m  5   n1

(

)

 1 1  0.071175894 =  2 − 2  n 5   1

7-20

n1 = 3 1 1 1  = 1.096776x107 m −1  2 − 2  c) λ 1 3  1 = 9.74912x106 m–1

(

)

λ

(

)  101 nmm  = 102.573 = 102.6 nm

λ = 1.02573x10−7 m  7.74

−9

Plan: Ionization occurs when the electron is completely removed from the atom, or when n final = ∞. We can use the equation for the energy of an electron transition to find the quantity of energy needed to remove completely the electron, called the ionization energy (IE). To obtain the ionization energy per mole of species, multiply by Avogadro’s number. The charge on the nucleus must affect the IE because a larger nucleus would exert a greater pull on the escaping electron. The Bohr equation applies to H and other one-electron species. Use the expression to determine the ionization energy of B4+. Then use the expression to find the energies of the hc transitions listed and use E = to convert energy to wavelength.

λ

Solution:  Z2  a) E = −2.18x10−18 J  2  n   

(

)

Z = atomic number

 1 1  2 ∆E = −2.18x10−18 J  2 − 2 Z n ninitial   final

(

)

 1 1  2  6.022x1023  ∆E = −2.18x10−18 J  2 − 2 Z   ∞ ninitial   1 mol   = (1.312796x106) Z2 for n = 1 b) In the ground state n = 1, the initial energy level for the single electron in B4+. Once ionized, n = ∞ is the final energy level.

Z = 5 for B4+. ∆E = IE = (1.312796x106) Z2 = (1.312796x106 J/mol)(52) = 3.28199x107 = 3.28x107 J/mol c) n final = ∞, n initial = 3, and Z = 2 for He+.  1 1  2 1  2 –19 −18  1 ∆E = −2.18x10−18 J  2 − 2  Z = −2.18x10 J  2 − 2  2 = 9.68889x10 J n n 3 ∞ initial     final hc E=

(

)

(

)

λ

(

(

)(

)

)

6.626x10−34 J•s 3.00x108 m/s hc λ (m) = = = 2.051628x10–7 m −19 E 9.68889x10 J  1 nm  λ (nm) = 2.051628x10−7 m  −9  = 205.1628 = 205 nm  10 m  d) n final = ∞, n initial = 2, and Z = 4 for Be3+.  1 1  2 1  2 –18 −18  1 ∆E = −2.18x10−18 J  2 − 2  Z = −2.18x10 J  2 − 2  4 = 8.72x10 J ∞ 2 n ∞ initial    

(

(

)

(

)

hc λ (m) = E

(6.626x10 =

−34

)(

J•s 3.00x108 m/s

8.72x10

−18

J

)

) = 2.279587x10 m –8

 1 nm  λ (nm) = 2.279587x10−8 m  −9  = 22.79587 = 22.8 nm  10 m 

(

)

7-21

7.75

a) Orbital D has the largest value of n, given that it is the largest orbital. b) l = 1 indicates a p orbital. Orbitals A and C are p orbitals. l = 2 indicates a d orbital. Orbitals B and D are d orbitals. c) In an atom, there would be four other orbitals with the same value of n and the same shape as orbital B. There would be two other orbitals with the same value of n and the same shape as orbital C. d) Orbital D has the highest energy and orbital C has the lowest energy.

7.76

Plan: Use the values and the equation given in the problem to calculate the appropriate values.

Solution: a) r n =

n2 h2 ε 0

π me e2

2 C2  12 6.626x10−34 J•s  8.854x10−12  J•m   kg•m 2 /s2   –11 –11 r1 =   = 5.2929377x10 = 5.293x10 m 2 −31 −19 J   π 9.109x10 kg 1.602x10 C

(

)

(

)(

)

2 C2  102 6.626x10−34 J•s  8.854x10−12  J•m   kg•m 2 /s2   –9 –9 b) r 10 =   = 5.2929377x10 = 5.293x10 m 2 −31 −19 J   π 9.109x10 kg 1.602x10 C

(

)

(

7.77

a) r n =

)

)(

n2 h2 ε 0

π me e2

2 C2  32 6.626x10−34 J•s  8.854x10−12  J•m   kg•m 2 /s2   –10 –10 r3 =   = 4.76364x10 = 4.764x10 m 2 −31 −19 J   π 9.109x10 kg 1.602x10 C

(

)

(

)(

)

b) Z = 1 for an H atom  Z2   12  E n = −2.18x10−18 J  2  = −2.18x10−18 J  2  = – 2.42222x10–19 = – 2.42x10–19 J n  3      c) Z = 3 for a Li atom  32   Z2  E n = −2.18x10−18 J  2  = −2.18x10−18 J  2  = – 2.18x10–1 3  n      d) The greater number of protons in the Li nucleus results in a greater interaction between the Li nucleus and its electrons. Thus, the energy of an electron in a particular orbital becomes more negative with increasing atomic number.

(

)

(

(

7.78

)

)

(

)

Plan: Refer to Chapter 6 for the calculation of the amount of heat energy absorbed by a substance from its specific heat capacity and temperature change (q = c x mass x ∆T). Using this equation, calculate the energy absorbed by the water. This energy equals the energy from the microwave photons. The energy of each photon can be calculated from its wavelength: E = hc/λ. Dividing the total energy by the energy of each photon gives the number of photons absorbed by the water.

Solution: q = c x mass x ∆T q = (4.184 J/g°C)(252 g)(98 – 20.)°C = 8.22407x104 J E=

hc

λ

=

( 6.626x10

−34

)(

J•s 3.00x108 m/s

1.55x10

−2

m

) = 1.28245x10 J/photon –23

7-22

(

7.79

)

1 photon   Number of photons = 8.22407x104 J  = 6.41278x1027 = 6.4x1027 photons −23   1.28245x10 J  One sample calculation will be done using the equation in the book: 3

3

−r  2 − r a0  1   1  2 − r a0  1   1 ψ=  =  = 1.465532x10–3 e a0  e   e   π   a0   π   52.92 pm  For r = 50 pm: −r

−50

ψ = 1.465532x10–3 e a0 = 1.465532x10–3 e 52.92 = 5.69724x10–4 ψ2 = (5.69724x10–4)2 = 3.24585x10–7 4πr2ψ2 = 4π(50)2(3.24585x10–7) = 1.0197x10–2 r (pm) ψ (pm–3/2) ψ2 (pm–3) 4πr2ψ2 (pm–1) –3 –6 0 1.47x10 2.15x10 0 50 0.570x10–3 0.325x10–6 1.02x10–2 –3 –6 100 0.221x10 0.0491x10 0.616x10–2 –3 –6 200 0.0335x10 0.00112x10 0.0563x10–2

The plots are similar to Figure 7.17A in the text.

7.80

Plan: In general, to test for overlap of the two series, compare the longest wavelength in the “n” series with the shortest wavelength in the “n+1” series. The longest wavelength in any series corresponds to the transition between the n 1 level and the next level above it; the shortest wavelength corresponds to the transition between the n 1  1 1 1  level and the n = ∞ level. Use the relationship = R  2 − 2  to calculate the wavelengths. n λ n2   1 Solution:  1  1 1 1  1  = R  2 − 2  = 1.096776x107 m −1  2 − 2     λ n2  n2   n1  n1 a) The overlap between the n 1 = 1 series and the n 1 = 2 series would occur between the longest wavelengths for n 1 = 1 and the shortest wavelengths for n 1 = 2.

Longest wavelength in n 1 = 1 series has n 2 equal to 2.

1 1  1 = 1.096776x107 m −1  2 − 2  = 8,225,820 m–1 λ 2  1 1 λ= = 1.215684272x10–7 = 1.215684x10–7 m −1 8,225,820 m Shortest wavelength in the n 1 = 2 series:

(

(

)

)

7-23

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1   1 –1 = 1.096776x107 m −1  2 −  = 2,741,940 m λ ∞2  2 1 λ= = 3.647052817x10–7 = 3.647053x10–7 m 2,741,940 m −1 Since the longest wavelength for n 1 = 1 series is shorter than shortest wavelength for n 1 = 2 series, there is no overlap between the two series. b) The overlap between the n 1 = 3 series and the n 1 = 4 series would occur between the longest wavelengths for n 1 = 3 and the shortest wavelengths for n 1 = 4.

Longest wavelength in n 1 = 3 series has n 2 equal to 4.

1 1   1 = 1.096776x107 m −1  2 − 2  = 533,155 m–1 λ 4  3 1 λ= = 1.875627163x10–6 = 1.875627x10–6 m 533,155 m −1 Shortest wavelength in n 1 = 4 series has n 2 = ∞.

1 1   1 = 1.096776x107 m −1  2 − = 685,485 m–1 2  λ 4 ∞   1 –6 λ= = 1.458821127x10 = 1.458821x10–6 m 685,485 m −1 Since the n 1 = 4 series shortest wavelength is shorter than the n 1 = 3 series longest wavelength, the series do overlap. c) Shortest wavelength in n 1 = 5 series has n 2 = ∞.

1 1   1 = 1.096776x107 m −1  2 − = 438,710.4 m–1 2  λ ∞  5 1 λ= = 2.27940801x10–6 = 2.279408x10–6 m −1 438,710.4 m Calculate the first few longest lines in the n 1 = 4 series to determine if any overlap with the shortest wavelength in the n 1 = 5 series: For n 1 = 4, n 2 = 5: 1 1   1 = 1.096776x107 m −1  2 − 2  = 246,774.6 m–1 λ 5  4 1 λ= = 4.052281x10–6 m 246,774.6 m −1 For n 1 = 4, n 2 = 6: 1 1   1 = 1.096776x107 m −1  2 − 2  = 380,825 m–1 λ 6  4 1 λ= = = 2.625878x10–6 m 380,825 m −1 For n 1 = 4, n 2 = 7: 1 1   1 = 1.096776x107 m −1  2 − 2  = 461,653.2 m–1 λ 7  4 1 λ= = 2.166128x10–6 m −1 461,653.2 m The wavelengths of the first two lines of the n 1 = 4 series are longer than the shortest wavelength in the n 1 = 5 series. Therefore, only the first two lines of the n 1 = 4 series overlap the n 1 = 5 series. d) At longer wavelengths (i.e., lower energies), there is increasing overlap between the lines from different series (i.e., with different n 1 values). The hydrogen spectrum becomes more complex, since the lines begin to merge into a more-or-less continuous band, and much more care is needed to interpret the information.

1

7-24

7.81

a) The highest frequency would correspond to the greatest energy difference. In this case, the greatest energy difference would be between E 3 and E 1 . ∆E = E 3 – E 1 = hν = (– 15x10–19 J) – (– 20x10–19 J) = 5x10–19 J ν = ∆E/h = (5x10–19 J)/(6.626x10–34 J•s) = 7.54603x1014 = 8x1014 s–1 λ = c/ν = (3.00x108 m/s)/(7.54603x1014 s–1) = 3.97560x10–7 = 4x10–7 m b) The ionization energy (IE) is the same as the reverse of E 1 . Thus, the value of the IE is 20x10–19 J/atom.

IE = (20x10–19 J/atom)(1kJ/103 J)(6.022x1023 atoms/mol) = 1204.4 = 1.2x103 kJ/mol c) The shortest wavelength would correspond to an electron moving from the n = 4 level to the highest level available in the problem (n = 6). ∆E = E 6 – E 4 = hc/λ = (– 2x10–19 J) – (– 11x10–19 J) = 9x10–19 J

λ = hc/∆E =

7.82

( 6.626x10

)(

−34

(

)

J•s 3.00x108 m/s  1 nm  2  −9  = 220.867 = 2x10 nm −19 9x10 J  10 m 

)

Plan: The energy differences sought may be determined by looking at the energy changes in steps. The hc wavelength is calculated from the relationship λ = .

E Solution: a) The difference between levels 3 and 2 (E 32 ) may be found by taking the difference in the energies for the 3 → 1 transition (E 31 ) and the 2 → 1 transition (E 21 ).

E 32 = E 31 – E 21 = (4.854x10–17 J) – (4.098x10–17 J) = 7.56x10–18 J

λ=

(

)(

)

6.626x10−34 J•s 3.00x108 m/s hc = = 2.629365x10–8 = 2.63x10–8 m E 7.56x10−18 J

(

)

b) The difference between levels 4 and 1 (E 41 ) may be found by adding the energies for the 4 → 2 transition (E 42 ) and the 2 → 1 transition (E 21 ).

E 41 = E 42 + E 21 = (1.024x10–17 J) + (4.098x10–17 J) = 5.122x10–17 J

λ=

(

)(

)

6.626x10−34 J•s 3.00x108 m/s hc = = 3.88091x10–9 = 3.881x10–9 m E 5.122x10−17 J

(

)

c) The difference between levels 5 and 4 (E 54 ) may be found by taking the difference in the energies for the 5 → 1 transition (E 51 ) and the 4 → 1 transition (see part b)).

E 54 = E 51 – E 41 = (5.242x10–17 J) – (5.122x10–17 J) = 1.2x10–18 J

λ=

(

)(

)

6.626x10−34 J•s 3.00x108 m/s hc = = 1.6565x10–7 = 1.66x10–7 m E 1.2x10−18 J

(

)

7.83

a) A dark green color implies that relatively few photons are being reflected from the leaf. A large fraction of the photons is being absorbed, particularly in the red region of the spectrum. A plant might adapt in this way when photons are in short supply — i.e., in conditions of low light intensity. b) An increase in the concentration of chlorophyll, the light-absorbing pigment, would lead to a darker green color (and vice versa).

7.84

Plan: For part a), use the equation for kinetic energy, E k = ½mu2. For part b), use the relationship E = hc/λ to find the energy of the photon absorbed. From that energy subtract the kinetic energy of the dislodged electron to obtain the work function.

Solution: a) The energy of the electron is a function of its speed leaving the surface of the metal. The mass of the electron is 9.109x10–31 kg.

2  J 1 1 9.109x10−31 kg 6.40x105 m/s  = 1.86552x10–19 = 1.87x10–19 J E k = mu2 = 2 2  2 2 kg•m /s  

(

)(

)

7-25

b) The minimum energy required to dislodge the electron (φ) is a function of the incident light. In this example, the incident light is higher than the threshold frequency, so the kinetic energy of the electron, E k , must be subtracted from the total energy of the incident light, hν, to yield the work function, φ. (The number of significant figures given in the wavelength requires more significant figures in the speed of light.)  10−9 m  λ (m) = ( 358.1 nm )  = 3.581x10–7 m  1nm   

( 6.626x10 J•s )( 2.9979x10 m/s ) = 5.447078x10 J ( 3.581x10 m ) −34

E = hc/λ =

8

–19

−7

Φ = hν – E k = (5.447078x10–19 J) – (1.86552x10–19 J) = 3.581558x10–19 = 3.58x10–19 J

 kg•m /s  (6.626x10 J•s ) a) λ = h/mu =   = 1.322568x10 m m   J   9.109x10 kg 5.5x10 ( )  s  −34

7.85

2

2

–8

−31

4

Smallest object = λ/2 = (1.322568x10–8 m)/2 = 6.61284x10–9 = 6.6x10–9 m

 kg•m /s  (6.626x10 J•s )   = 2.424708x10 m m   J   (9.109x10 kg )  3.0x10 s  −34

b) λ = h/mu =

2

2

–11

−31

7

Smallest object = λ/2 = (2.424708x10–11 m)/2 = 1.212354x10–11 = 1.2x10–11 m 7.86

Plan: Examine Figure 7.3 and match the given wavelengths to their colors. For each salt, convert the mass of salt to moles and multiply by Avogadro’s number to find the number of photons emitted by that amount of salt hc (assuming that each atom undergoes one-electron transition). Use the relationship E = to find the energy of

λ

one photon and multiply by the total number of photons for the total energy of emission.

Solution: a) Figure 7.3 indicates that the 641 nm wavelength of Sr falls in the red region and the 493 nm wavelength of Ba falls in the green region. b) SrCl 2  1 mol SrCl 2   6.022x1023 photons  22 Number of photons = ( 5.00 g SrCl 2 )   = 1.8994449x10 photons   1 mol SrCl 2  158.52 g SrCl 2     10−9 m  = 6.41x10–7 m  1nm   

λ (m) = ( 641 nm ) 

( 6.626x10 =

−34

)(

)

J•s 3.00x108 m/s  1 kJ  –22 E photon =  3  = 3.10109x10 kJ/photon λ 6.41x10−7 m 10 J   −22   3.10109x10 kJ E total = 1.8994449x1022 photons   = 5.89035 = 5.89 kJ  1 photon   BaCl 2  1 mol BaCl 2   6.022x1023 photons  = 1.44620557x1022 photons Number of photons = ( 5.00 g BaCl 2 )     1 mol BaCl 2  208.2 g BaCl 2   

hc

(

)

 10−9 m  = 4.93x10–7 m  1nm   

λ (m) = ( 493 nm ) 

7-26

( 6.626x10

−34

)(

)

J•s 3.00x108 m/s  1 kJ  –22  3  = 4.0320487x10 kJ/photon −7 λ 4.93x10 m  10 J   4.0320487x10−22 kJ  E total = 1.44620557x1022 photons   = 5.83117 = 5.83 kJ  1 photon   7.87 a) The highest energy line corresponds to the shortest wavelength. The shortest wavelength line is given by  1  1 1 1  1  = R  2 − 2  = 1.096776x107 m −1  2 − 2  n n λ n2  n2   1  1 E photon =

hc

=

(

)

(

)

 1 nm  1 1  −1  1 7  −9  = 1.096776x10 m  2 −  3282 nm  10 m  ∞ 2   n1 304,692 m–1 = (1.096776x107 m–1) (1/n2) 1/n2 = 0.0277807 n=6 b) The lowest energy line corresponds to the longest wavelength. The longest wavelength line is given by  1  1 1  = 1.096776x107 m −1  2 − 2   n1 λ n + 1 ( ) 1    1  1  1 nm  1  = 1.096776x107 m −1  2 −   −9 2 7460 nm  10 m  n ( n1 + 1)   1  1  1  134,048 m–1 = 1.096776x107 m −1  2 − 2   n1 n + 1 ( ) 1    1  1  0.0122220 =  2 − 2  n1  n + 1 ( ) 1  

)

(

(

)

(

(

)

)

Rearranging and solving this equation for n 1 yields n 1 = 5. (You and your students may well need to resort to trialand-error solution of this equation!)

7.88

Plan: Examine Figure 7.3 to find the region of the electromagnetic spectrum in which the wavelength lies.

Compare the absorbance of the given concentration of Vitamin A to the absorbance of the given amount of fishliver oil to find the concentration of Vitamin A in the oil.

Solution: a) At this wavelength the sensitivity to absorbance of light by Vitamin A is maximized while minimizing interference due to the absorbance of light by other substances in the fish-liver oil. b) The wavelength 329 nm lies in the ultraviolet region of the electromagnetic spectrum. c) A known quantity of vitamin A (1.67x10–3 g) is dissolved in a known volume of solvent (250. mL) to give a standard concentration with a known response (1.018 units). This can be used to find the unknown quantity of Vitamin A that gives a response of 0.724 units. An equality can be made between the two concentration-toabsorbance ratios.  1.67x10−3 g  = 6.68x10–6 g/mL Vitamin A Concentration (C 1 , g/mL) of Vitamin A =   250. mL    Absorbance (A 1 ) of Vitamin A = 1.018 units.

Absorbance (A 2 ) of fish-liver oil = 0.724 units Concentration (g/mL) of Vitamin A in fish-liver oil sample = C 2

A1 A = 2 C1 C2

7-27

C2 =

(

( 0.724 ) 6.68x10 A2C1 = A1 (1.018)

−6

g/mL

) = 4.7508x10 g/mL Vitamin A –6

 4.7508x10−6 g Vitamin A  –3 Mass (g) of Vitamin A in oil sample = ( 500. mL oil )   = 2.3754x10 g Vitamin A  1 mL oil  

( 2.3754x10 g ) = 1.92808x10 = 1.93x10 g Vitamin A/g oil −3

Concentration of Vitamin A in oil sample =

λ = hc/E =

–2

( 6.626x10 J•s )( 3.00x10 m/s )  1 nm  = 261.897 = 262 nm    10 m  ( 7.59x10 J ) −34

7.89

–2

( 0.1232 g Oil )

8

−9

−19

Silver is not a good choice for a photocell that uses visible light because 262 nm is in the ultraviolet region.

7.90

Mr. Green must be in the dining room where green light (520 nm) is reflected. Lower frequency, longer wavelength light is reflected in the lounge and study. Both yellow and red light have longer wavelengths than green light. Therefore, Col. Mustard and Ms. Scarlet must be in either the lounge or study. The shortest wavelengths are violet. Prof. Plum must be in the library. Ms. Peacock must be the murderer.

7.91

Ek = v=

1 mv2 2

Ek 1 m 2

=

4.71x10−15 J

 kg•m 2 /s2  8   = 1.01692775x10 m/s 1 −31 J 9.109x10 kg   2

(

)

 kg•m /s  ( 6.626x10 J•s )   = 7.15304x10 = 7.15x10 m m   J   9.109x10 kg 1.01692775x10 ( )  s  −34

λ = h/mv =

7.92

2

2

–12

−31

–12

8

Plan: First find the energy in joules from the light that shines on the text. Each watt is one joule/s for a total hc of 75 J; take 5% of that amount of joules and then 10% of that amount. Use E = to find the energy of one

λ

photon of light with a wavelength of 550 nm. Divide the energy that shines on the text by the energy of one photon to obtain the number of photons.

Solution: The amount of energy is calculated from the wavelength of light:  10−9 m  λ (m) = ( 550 nm )  = 5.50x10–7 m  1 nm    E=

hc

λ

(6.626x10 =

−34

)(

J•s 3.00x108 m/s −7

5.50x10 m

) = 3.614182x10 J/photon –19

 1 J/s  Amount of power from the bulb = ( 75 W )   = 75 J/s 1W   5%  Amount of power converted to light = ( 75 J/s )   = 3.75 Js  100% 

 10%  Amount of light shining on book = ( 3.75 J/s )   = 0.375 J/s  100%  1 photon  0.375 J    = 1.0376x1018 = 1.0x1018 photons/s Number of photons:   −19  s 3.614182x10 J   

7-28

7.93

a) Sodium ions emit yellow-orange light, and potassium ions emit violet light. b) The cobalt glass filter absorbs the yellow-orange light, while the violet light passes through. c) Sodium salts used as the oxidizing agents would emit intense yellow-orange light which would obscure the light emitted by other salts in the fireworks.

7.94

a) 6CO 2 (g) + 6H 2 O(l) → C 6 H 12 O 6 (s) + 6O 2 (g) ∆H rxn = {(1 mol) ∆H f [C 6 H 12 O 6 ] + (6 mol) ∆H f [O 2 ]} – {(6 mol) ∆H f [CO 2 ] + (6 mol) ∆H f [H 2 O]} ∆H rxn = [–1273.3 kJ + 6(0.0 kJ)] – [6(–393.5 kJ) + 6(–285.840 kJ)] = 2802.74 = 2802.7 kJ ∆H rxn = 2802.7 kJ (for 1.00 mol C 6 H 12 O 6 ) 6CO 2 (g) + 6H 2 O(l) → C 6 H 12 O 6 (s) + 6O 2 (g)

( 6.626x10

−34

)(

)

J•s 3.00x108 m/s  1 nm  –19  −9  = 2.9232353x10 J/photon 680. nm 10 m   3  10 J    1 photon Number of photons = ( 2802.7 kJ )  = 9.5877x1024 = 9.59x1024 photons  1 kJ   2.9232353x10−19 J     

b) E = hc/λ =

7.95

Plan: In the visible series with n final = 2, the transitions will end in either the 2s or 2p orbitals since those are the only two types of orbitals in the second main energy level. With the restriction that the angular momentum quantum number can change by only ±1, the allowable transitions are from a p orbital to 2s (l = 1 to l = 0), from an s orbital to 2p (l = 0 to l = 1), and from a d orbital to 2p (l = 2 to l = 1). The problem specifies a change in energy level, so n init must be 3, 4, 5, etc. (Although a change from 2p to 2s would result in a +1 change in l, this is not a change in energy level.)

Solution: The first four transitions are as follows: 3s → 2p 3d → 2p 4s → 2p 3p → 2s

7.96

a)

0.9

Absorbance

0.7 0.5 0.3 0.1 0 0

10 20 Concentration, mol/L

A = kC k = 4.5x103 M–1 = slope b) A/k = C = (0.55)/(4.5x103 M–1) = 1.2222x10–4 = 1.2x10–4 M

7-29

−5

x 10