Answer key to homework for chapters 46,47

Answer key to homework for chapters 46,47
Chapter 46
E46-2 (a) λ = h/p and p2/2m = K, then
(b) K = eV , so
E46-15 First find the wavelength of the neutrons. For non-relativistic particles λ= h/p and p2/2m = K, so
Bragg reflection occurs when 2d sinθ = λ, so
θ = arcsin (14 pm)/2(73.2 pm) = 5.5o
E46-25 We want v ≈∆v, which means p ≈∆p. Apply Eq. 46-8, and
According to Eq. 46-1, the de Broglie wavelength is related to the momentum by
λ=h/p, so
E46-30 The wave number, k, is given by
(a) For the proton mc2 = 938 MeV, so
A current of 1 kA corresponds to
protons per seconds. The time required for one proton to pass is then
That's 10104 years!
Chapter 47
E47-7 (a) We will use the result of Exercise 45-1 for hc .
(b) The magnitude of the momentum is exactly known because E = p 2/ 2m . This momentum is given by
3
19
23
8
94.410 1.610
5.0310
kg m/s 310
p --⨯⨯⨯=
=⨯⋅⨯
It can be also get from another way,
34
23
12
15 6.6310 5.0510
kg m /s
2298.510
nh p L
---⨯⨯=
=
=
=
=⨯⋅⨯⨯
What we don't know is in which direction the particle is moving. It is bouncing back and forth between the walls of the box, the uncertainty in the momentum is then ∆p=2p =1.01⨯10-22 kg .m/s
(c) The uncertainty in the position is 98.5 pm; the electron could be anywhere inside the well.
E47-8 The probability distribution function is
We want to integrate over the central third, or
E47-15 The binding energy is the energy required to remove the electron. If the energy of the electron is negative, then that negative energy is a measure of the energy required to set the electron free.
The first excited state is when n = 2 in Eq. 47-21. It is not necessary to re-evaluate the constants in this equation every time, instead, we start from
Then the first excited state has energy
The binding energy is then 3.4 eV.
E47-33 (a) We'll use Eqs. 47-25 and 47-26. At r = 0。