Type:
Instructor Manual
Resource:
Modern Physics
Edition:
4th Edition
Author(s):
Kenneth S. Krane
Chapter 1 1. (a) Conservation of momentum gives px ,initial = px ,final , or mH vH,initial + mHe vHe,initial = mH vH,final + mHe vHe,final
Solving for vHe,final with vHe,initial = 0 , we obtain
vHe,final =
mH (vH,initial − vH,final )
mHe (1.674 ×10−27 kg)[1.1250 ×107 m/s − (−6.724 ×106 m/s)] = = 4.527 ×106 m/s −27 6.646 ×10 kg
(b) Kinetic energy is the only form of energy we need to consider in this elastic collision. Conservation of energy then gives K initial = K final , or 1 2
2 2 2 2 mH vH,initial + 12 mHe vHe,initial =12 mH vH,final + 12 mHe vHe,final
Solving for vHe,final with vHe,initial = 0 , we obtain
vHe,final = =
2 2 − vH,final mH (vH,initial )
mHe (1.674 ×10−27 kg)[(1.1250 ×107 m/s) 2 − (−6.724 ×106 m/s) 2 ] = 4.527 ×106 m/s −27 6.646 ×10 kg
2. (a) Let the helium initially move in the x direction. Then conservation of momentum gives: : = px ,initial px ,final = mHe vHe,initial mHe vHe,final cos θ He + mO vO,final cos θ O p y ,initial p= 0 mHe vHe,final sin θ He + mO vO,final sin θ O = y ,final : From the second equation,
sin θ He m v (6.6465 ×10−27 kg)(6.636 ×106 m/s)(sin 84.7°) vO,final = 2.551×106 m/s − He He,final = − = (2.6560 ×10−26 kg)[sin(−40.4°)] mO sin θ O (b) From the first momentum equation,
Chapter 1
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vHe,initial =
mHe vHe,final cos θ He + mO vO,final cos θ O
mHe (6.6465 × 10 kg)(6.636 × 106 m/s)(cos 84.7°) + (2.6560 × 10−26 kg)(2.551 × 106 m/s)[cos(−40.4°)] = 6.6465 × 10−27 kg = 8.376 × 106 m/s −27
3. (a) Using conservation of momentum for this one-dimensional situation, we have px ,initial = px ,final , or
mHe vHe + mN vN = mD vD + mO vO Solving for vO with vN = 0 , we obtain
mHe vHe − mD vD (3.016 u)(6.346 ×106 m/s) − (2.014 u)(1.531×107 m/s) vO = = = −7.79 ×105 m/s mO 15.003 u (b) The kinetic energies are: 2 1.008 ×10−13 J K initial =12 mHe vHe + 12 mN vN2 =12 (3.016 u)(1.6605 ×10−27 kg/u)(6.346 ×106 m/s) 2 = K final = 12 mD vD2 + 12 mO vO2 = 12 (2.014 u)(1.6605 ×10−27 kg/u)(1.531×107 m/s) 2 3.995 ×10−13 J + 12 (15.003 u)(1.6605 ×10−27 kg/u)(7.79 ×105 m/s) 2 =
As in Example 1.2, this is also a case in which nuclear energy turns into kinetic energy. The gain in kinetic energy is exactly equal to the loss in nuclear energy.
4. Let the two helium atoms move in opposite directions along the x axis with speeds v1 and v2 . Conservation of momentum along the x direction ( p x ,initial = p x ,final ) gives
0= m1v1 − m2 v2 or
v1 = v2
The energy released is in the form of the total kinetic energy of the two helium atoms:
K1 + K 2 = 92.2 keV Because v1 = v2 , it follows that K= K= 46.1 keV , so 1 2 = v
2 K1 = m1
2(46.1 ×103 eV)(1.602 ×10−19 J/eV) = 1.49 ×106 m/s (4.00 u)(1.6605 ×10−27 kg/u)
v2= v= 1.49 ×106 m/s 1
Chapter 1
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5. (a) The kinetic energy of the electrons is −31 2 1 1 Ki = kg)(1.76 ×106 m/s) = 14.11×10−19 J 2 mvi = 2 (9.11× 10
In passing through a potential difference of ∆V = Vf − Vi = +4.15 volts , the potential energy of the electrons changes by
∆U =∆ q V =− ( 1.602 ×10−19 C)( + 4.15 V) = −6.65 ×10−19 J Conservation of energy gives K i + U i = K f + U f , so
K= K i + (U i − U f = ) K i − ∆U= 14.11×10−19 J + 6.65 ×10−19 = J 20.76 ×10−19 J f 2Kf = m
= vf
2(20.76 ×10−19 J) = 2.13 ×106 m/s −31 9.11× 10 kg
(b) In this case ∆V = −4.15 volts, so ∆U = +6.65 × 10−19 J and thus
K= K i − ∆U= 14.11×10−19 J − 6.65 ×10−19 = J 7.46 ×10−19 J f = vf
2Kf = m
2(7.46 ×10−19 J) = 1.28 ×106 m/s −31 9.11×10 kg
6. (a) ∆x A =v∆t A =(0.624)(2.997 ×108 m/s)(124 ×10−9 s) =23.2 m (b) ∆xB =v∆t B =(0.624)(2.997 ×108 m/s)(159 ×10−9 s) =29.7 m 7. With T = 35°C = 308 K and= P 1.22 atm = 1.23 ×105 Pa ,
N P 1.23 × 105 Pa = = = 2.89 × 1025 atoms/m3 -23 V kT (1.38 × 10 J/K)(308 K) so the volume available to each atom is (2.89 × 1025/m3)−1 = 3.46 × 10−26 m3. For a spherical atom, the volume would be 4 3
−10 4 π R3 = m)3 =× 1.50 10−30 m3 3 π (0.710 × 10
The fraction is then
1.50 ×10−30 = 4.34 ×10−5 −26 3.46 ×10 Chapter 1
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8. Differentiating N(E) from Equation 1.23, we obtain
dN 2 N 1 1 −1/ 2 − E / kT 1 − E / kT = E e + E1/ 2 − e 3/ 2 2 dE π (kT ) kT To find the maximum, we set this function equal to zero: 1 1 E E −1/ 2 e − E / kT − 0 = 3/ 2 π (kT ) 2 kT
2N
Solving, we find the maximum occurs at E = 12 kT . Note that E = 0 and E = ∞ also satisfy the equation, but these solutions give minima rather than maxima. 9. For this case kT= (280 K)(8.617 ×10−5 eV/K) = 0.0241 eV . We take dE as the width of the interval (0.0012 eV) and E as its midpoint (0.0306 eV). Then dN = N ( E ) dE =
10.
2N
1 (0.0306 eV)1/2 e − (0.0306 eV)/(0.0241 eV) (0.0012 eV) = 1.8 ×10−2 N 3/2 π (0.0241 eV)
(a) From Eq. 1.33, ∆Eint=
5 2
nR ∆T=
5 2
(2.37 moles)(8.315 J/mol ⋅ K)(65.2 K)= 3.21×103 J
7 2
(2.37 moles)(8.315 J/mol ⋅ K)(65.2 K)= 4.50 ×103 J
(b) From Eq. 1.34, ∆Eint=
7 2
nR ∆T=
(c) For both cases, the change in the translational part of the kinetic energy is given by Eq. 1.31: ∆Eint=
11.
3 2
nR ∆T=
5 2
(2.37 moles)(8.315 J/mol ⋅ K)(65.2 K)= 1.93 ×103 J
After the collision, m 1 moves with speed v1′ (in the y direction) and m 2 with speed v2′ (at an angle θ with the x axis). Conservation of energy then gives E initial = E final : 1 2
1 ′2 1 ′2 m1v12 = 2 m1v1 + 2 m2 v2
or
v2 = v1′2 + 3v2′2
Conservation of momentum gives:
Chapter 1
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