Key 6
1. Key to Prob. 4.2
We use the simpler classical expression p = (2mK)1/2 instead of the relativistic
expression p = K(1 + 2mc2 /K)1/2 /c. 2mc2 for electrons is approximately equal
to 1 MeV and for our cases, K << 1 MeV.
(a) K = 50 eV.
λ = hc/[2(0.511 MeV/c2 )(50 eV)]1/2 = 0.173 nm.
(b) K = 50 keV.
λ = hc/[2(0.511 MeV/c2 )(50 × 103 eV)]1/2 = 5.49 × 10−3 nm.
If we use the relativistic expression we will get λ = 0.00536 nm.
2. Key to Prob. 4.9
m = 0.20 kg and mgh = mv 2 /2 ⇒ v = (2gh)1/2 . So,
q
p = mv = m 2gh = (0.20)[2(9.80)(50)]1/2 = 6.261 kg. m/s.
λ = h/p = (6.626 × 10−34 Js)/6.261 kgm/s = 1.06 × 10−34 m.
3. Key to Prob. 4.22
4p = m4v = (0.05kg)(10−3 × 30m/s) = 1.5 × 10−3 kg.m/s. Therefore, 4x =
h̄/24p = 3.51 × 10−32 m.
4. Key to Prob. 4.25
The woman tries to hold the pellets just above the spot of the floor, with the
same x-coordinate. In doing so, she gives them some x-velocity, atleast as large
1
as in 4xm4vx = h̄/2. In falling they will scatter to fill an angle θ around
the vertical, with m4vx = p sin θ = mv4x/H where v is the final speed. SO
eliminating 4va we get,
4xmv4x/H = h̄/2 ⇒ (4x)2 = h̄H/(2mv).
We also have H = gt2 /2 and v = gt, so v = (2gH)1/2 . Then,
(4x)2 = (h̄/2m)(H/2g)1/2 ⇒ 4x = (h̄/2m)1/2 (H/2g)1/4 .
(b) 4x = 1.84 × 10−16 m.
5. Key to Prob. 4.28
(a) 4x4p = h̄ so if 4x = r, 4p ≈ h̄/r.
(b) K = p2 /2me ≈ (4p)2 /2me = h̄2 /(2me r2 ).
U = −ke2 /r. So, E = h̄2 /2me r2 − ke2 /r.
(c) To minimize E, take
dE/dr = −h̄2 /me r3 + ke2 /r2 = 0 ⇒ r = h̄2 /me ke2 = Bohr radius = a0 .
Then,
E = (h̄2 /2me )(me ke2 /h̄2 )2 − ke2 (me ke2 /h̄2 ) = me k 2 e4 /2h̄2 = −13.6eV.
6. Key to Prob. 4.29
To find the energy width of the γ-ray we use, 4E4t ≥ h̄/2 or,
4E ≥ h̄/(24t) ≥ 6.58 × 10−16 eV s/[(2)(0.10 × 10−9 s)] ≥ 3.29 × 10−6 eV.
Since the intrinsic width is so much less than the experimental resolution of ±5
eV, it can’t be measured with this method.
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