PHYS 212-071
HW-6 Solutions
(Numbers refer to 2nd Edition)
5. To “observe” small objects, one measures the diffraction of particles whose de Broglie
wavelength is approximately equal to the object’s size. Find the kinetic energy (in
electron volts) required for electrons to resolve (a) a large organic molecule of size 10
nm, (b) atomic features of size 0.10 nm, and (c) a nucleus of size 10 fm. Repeat these
calculations using alpha particles in place of electrons.
p
h

 pc 
hc

hc /  
p 2  pc 


2
2m 2mc
2mc 2
2
Nonrelativistic: KE 
Relativistic: KE  E  mc 2 
2
 pc2  mc 2  mc2
hc  1240.8eV .nm
Electrons
me c 2  0.511  10 6 eV
(a)   10nm  KE  0.015eV
(b)   0.1nm  KE  150eV
(c)   10 fm  KE  123.6MeV
Alpha Particles
m c 2  3728.4  10 6 eV
(a)   10nm  KE  2eV
(b)   0.1nm  KE  20meV
(c)   10 fm  KE  2.07 MeV
1
24. We wish to measure simultaneously the wavelength and position of a photon. Assume
that the wavelength measurement gives   6000 Angstroms with an accuracy of one
part in one million, i.e.,


 10 6 . What is the minimum uncertainty in the position of
the photon?
p
h


xp 
p

 h 
  
 p  p 
 
p


 




 x 
 x 
 x 
2
2p
2 /    h /  
4  /  

  6000 A,


 10 6  x min  4.8cm
29. An excited nucleus with a lifetime of 0.100 ns emits a  ray of energy 2.00 MeV.
Can the energy width (uncertainty in energy, E ) of this 2.00 MeV  emission line be
directly measured if the best  detectors can measure energies to  5 eV?
  0.100ns
E  2.00MeV
E exp  5eV

 6.582  10 16 eV .s
E width     E width 

 E width  3.3  10 8 eV  Eexp
9
2
2
2  0.1  10 s
The energy width cannot be resolved.
2
34. Robert Hofstadter won the 1961 Nobel Prize in Physics for his pioneering work in
scattering 20-GeV electrons from nuclei. (a) What is the  factor for a 20-GeV electron,
 v2
where   1  2
 c



1
2
. What is the momentum of the electron in kg.m/s? (b) What is the
wavelength of a 20-GeV electron and how does it compare to the size of a nucleus?
 
E
mc 2
E  20GeV  20  10 3 MeV
mc 2  0.511MeV
(a)   39139
(b) pc  20GeV  p 
(c)  
20  10 9  1.6  10 19 J
 p  10 17 kg.m / s
8
3  10 m / s
h hc 1240.8eVnm


   0.062 fm  Nuclear size.
p pc 20  10 9 eV
3