King Fahd University of Petroleum and Minerals
Department of Physics
PHYS 212-MODERN PHYSICS
Term 071
First Major Exam
SOLUTION
Wednesday 07 November 2007
08:00-08:50
Name:
ID No.:
1
Q1. The muon is an unstable particle that spontaneously decays into an electron and
two neutrinos. If the number of muons at t  0 is N 0 , the number at time t is given
by N  N 0 e  t /  , where  is the mean lifetime, equal to 2.2 s . Suppose the muons
move at a speed of 0.95 c and there are 5.0  10 4 muons at t  0 . (a) What is the
observed lifetime of the muons? (b) How many muons remain after traveling a
distance of 3.0 km ?
A1. In the rest frame of the muon we have N  N 0 e
t

respect to which the muon is moving we have N  N 0 e
 v2
v  0.95c , and   1  2
 c
N  N0e

d
v
 5  10  e
4




1
2
. In a reference frame with
t  

, where t   d ,     ,
v
 3.2 ; therefore
3000
3.20.953108 2.210 6
 11,210 muons.
Q2. A radium isotope decays to a radon isotope
helium nucleus) according to the decay scheme
222Rn
226Ra
by emitting an  particle (a
 222Rn + 4He. The masses of
the atoms are 226.0254 (Ra), 222.0175 (Rn), and 4.0026 (He). How much energy is
released as a result of this decay?
A2. In the decay
226
Ra  222Rn  4He
the conservation of energy in the rest frame of the
226
Ra nucleus can be written as
M Ra c 2  M Rn c 2  K Rn  M He  K He
where K is the kinetic energy. Therefore, the energy released is
K Rn  K He  ( M Ra  M Rn  M He )c 2  (226.0254u  222.0175u  4.0026u )c 2  5  10 3 uc 2
MeV
We also know that 1u  931.5 2 . Therefore the energy released in the decay is
c
5  10 3  931.5  4.94MeV
2
Q3. Photons of wavelength 450 nm are incident on a metal. The most energetic
electrons ejected from the metal are bent into a circular arc of radius 20 cm by a
magnetic field whose strength is equal to 2.0  10 5 T. What is the work function of
the metal?
A3. We have
p  qBR  pc  qBRc  1.6  10 19 C  2.0  10 5 T  0.2m  3  108 m / s  1.2  10 3 eV  1.2keV.
We also have
E 2  ( pc) 2  (mc 2 ) 2  (mc 2  K ) 2  K  ( pc) 2  (mc 2 ) 2  mc 2
Since mc 2  511keV , we get K  1.41eV
From the equation
K  h      h  K 
hc

K,
we get

1240eVnm
 1.41nm    2.76eV
450nm
Q4. X-rays of wavelength 0.200 nm are scattered from a block of carbon. If the
scattered radiation is detected at 90o to the incident beam, find (a) the Compton
shift  and (b) the kinetic energy imparted to the electron.
A5. We have   0.00243nm(1  cos  )  0.00243nm for   90 0 . Therefore the
wavelength of the scattered photon is
       0.200nm  0.00243nm  0.20243nm
The kinetic energy of the electron is
1 1
K  h  h   hc(  )  74.5eV
 
3
Q5. (a) Calculate the longest and shortest wavelengths of the Balmer series. (b)
Determine the photon energies (in eV) corresponding to these wavelengths.
A5. (a) We have the Rydberg formula
 1
1 
 R 2  2  ,
n


 f ni 
1
Where R  1.097  10 7 m 1 is the Rydberg constant. For the Balmer series n f  2 .
The longest wavelength corresponds to ni  3 , and the shortest wavelength
corresponds to ni   . Substituting for these in the above formula we get
longest  656.3nm and shortest  364.6nm .
(b) The energies can be found by using the relation E  h 
hc

. After substitution
for the wavelengths found in (a) we get Elongest  1.9eV and E shortest  3.4eV .
Q6. Explain why Emission spectra contain more spectral lines than absorption
spectra for the same element.
A6. In emission spectra the higher energy electron can cascade down to many lower
energy orbits therefore giving rise to many emission spectral lines. In absorption
spectra, however, the lower energy electron can move up to only ONE higher energy
orbit thereby creating fewer spectral lines.
Q7. What is the single most important experimental observation supporting the
explanation of the photoelectric effect in terms of the particle nature of light.
A7. The fact that the stopping voltage (maximum kinetic energy of photoelectrons)
does not depend on the intensity of the light but depends ONLY on its frequency.
4
Q8. Explain why we believe that there are tiny nuclei inside atoms.
A8. Because of the large number of alpha particles scattered at large angles in
Rutherford scattering.
Q9. What is the postulate due to Bohr that is necessary to derive the discreteness of
atomic radii?
A9. L  n , where L is the angular momentum of the electron moving around the
nucleus.
Q10. Explain why it is necessary to use Lorentz transformations instead of Galilean
transformations when we want to relate events in a moving reference frame to
events in another, fixed reference frame?
A10. Because they are the only transformations that leave the speed of light constant
(as observed experimentally) between these reference frames.
5