PH300
Fall 2013
HW7 answer key
1) Find the de Broglie wavelength for each of the following electrons
with the specified kinetic energy:
a) electron of 20 keV in a TV tube,
b) conduction electron of 5.4 eV in a metal,
c) orbiting electron of 13.6 eV in a hydrogen atom,
d) orbiting electron of 91 keV in a lead atom.
Non-relativistic de Broglie wavelength:
momentum p = mv, and kinetic energy is KE = ½ mv2 using these
equations and doing some algebra v = (2*KE*m)1/2
The de Broglie wavelength is λ = h/p = h*c/(p*c) = h*c/(2*KE*m*c2)1/2
a) λ = 1240 eV*nm/(2*20 keV*0.511 MeVc-2*c2)1/2 = 0.00867 nm
b) λ = 1240 eV*nm/(2*5.4 eV*0.511 MeVc-2*c2)1/2 = 0.5278 nm
c) λ = 1240 eV*nm/(2*13.6 eV*0.511 MeVc-2*c2)1/2 = 0.3326 nm
d) λ = 1240 eV*nm/(2*91 keV*0.511 MeVc-2*c2)1/2 = 0.00407 nm
Relatavistic de Broglie wavelength:
momentum p = γmv, and kinetic energy KE = (γ -1)mc2
using KE γ can be determined: γ = 1 + KE/mc2
and from γ velocity can be determined: v = (1 - γ-2)1/2*c
The de Broglie wavelength is: λ = h/p = h*c/(p*c) = h*c/(γmvc)
a) λ = 1240 eV*nm/(1.03914*0.511 MeVc-2*(0.27187c)*c)
= 0.00859 nm
b) λ = 1240 eV*nm/(1.0001*0.511 MeVc-2*(0.004597c)*c)
= 0.5278 nm
c) λ = 1240 eV*nm/(1.00003*0.511 MeVc-2*(0.007296c)*c)
= 0.3326 nm
d) λ = 1240 eV*nm/(1.1781*0.511 MeVc-2*(0.52865c)*c)
= 0.003896 nm
2) An electron initially at rest is accelerated across a potential
difference of 3000 volts. What are its wavelength, momentum, kinetic
energy, and total energy?
This electron travels across a 3 kV potential. This means the energy the electron
has is 3 keV of kinetic energy.
The total energy is Etot = mc2 + KE = 5.14x105 eV
The momentum can be determined by E2 = p2c2 + m2c4
p = (E2 – m2c4)1/2*c-1
= 5.55x104 eV/c
The de Broglie wavelength: λ = hc/pc = 1240 eV*nm/5.55x104 eV = 0.0224 nm
3) The nucleus of the aluminum atom has a diameter of 7.2x10-15 m.
Consider one of the protons in this nucleus. The uncertainty in the
position of this proton is necessarily less than 7.2x10-15 m. What is
the minimum uncertainty in its velocity and momentum?
Using the Heisenberg’s uncertainty principal
∆x∆p ≥ /2 → ∆p = /(∆x*2) = (1.0546x10-34 J*s)/(2*7.2x10-15 m)
= 7.3236x10-21 kg*m/s
∆p = m∆v, where m = 1.6726x10-27 kg
∆v = ∆p/m = (7.3236x10-21 kg*m)/(1.6726x10-27 kg*m/s)
= 4.3786x106 m/s or 0.0146c
4) In an electron microscope we wish to study particles of diameter about 0.10
μm (about 1000 times
the size of a single atom). (a) What should be the de Broglie wavelength of the
electrons? (b) Through
what potential difference should the electrons be accelerated to have that de
Broglie wavelength?
(a)The best wavelength, which is used to view an object, is the order of the size
of that object. So de broglie wavelength λ=1*10^-7m
(b)Momentum p=h/ λ=6.63*10-27
Thus kinetic energy k=p^2/(2 Me)=2.41*10^-23J=1.51*10^-4ev
Then potential should be 1.51*10^-4volts
5) Krane Ch5 #8
6) For the infinite square-well potential, find the probability that a
particle in its ground state is in each third of the one-dimensional box
(of length L) (T&R problem 6-17).
The probability of finding a particle is determined by integrating the probability
density over the region of interest. There are three regions of interest
a) 0 to L/3, b) L/3 to 2L/3 and c) 2L/3 to L
The probability density for an infinite square well potential in the ground state is g
ψ
2
=
2
 πx 
s i 2n
 and the integral of this is 2/L{x/2-L/π4*sin(πx/L)}
L
 L 
region a)
∫
L
0
/ 3

2L
L
π2
ψ 2 d = x  −
s  i n − (0 − 0) 
L 6 π4
 3 

= 2/L(L/6 – (L/π)*0.2165) = 1/3 – (2/π)*0.2165
region b)
∫
2
L
ψ2d =
L
/
3
/ 2 3
L
L
L
 π 2  
 π 4   L
s  i   n−  −
s  i  n
x −
L  3 π 4  3    6 π 4  3  
= 2/L(L/6 + (L/π)*0.433) = 1/3 + (2/π)*0.433
region c)
∫
L
2
ψ2d =
L
/
2  L
L
x −
s
3
L  2 π 4
(π )i − n L −

3
L
 π 4  
s  i  n
π 4  3  
= 2/L(L/6 - (L/π)*0.2165) = 1/3 – (2/π)*0.2165
One way to check this is to find out of all parts add up to 1 (the particle has to be
some where in the well)
a) + b) + c) = 1/3 + 1/3 + 1/3 + (2/π)*(-0.2165 + 0.433 – 0.2165)
= 1 + (2/π)*0 = 1 the answer checks out
Extra Credit
Krane 4-20
From figure 4.28, we can get ∆E=120Mev. Using uncertainty relationship ∆E*
∆t= ћ.
Then ∆t= ћ/∆E=5.48*10^-24S