Clayton State University
Department of Natural Sciences
September 1, 2006
Physics 3650 – Quiz 1
Name __SOLUTION___________________________________
1. You are a passenger on a spaceship. As the speed of the spaceship increases, you
would observe
a. your watch slowing down.
b. your watch speeding up.
c. your watch losing time.
d. nothing unusual about your watch.
2. A spaceship carrying a light clock moves at a speed of 0.960c relative to an observer
on Earth. Relative to a frame of reference that is fixed with respect to Earth, how long
does it take the clock to advance 1.00 s?
a. 1.22 s.
b. 2.59 s.
c. 3.57 s. (See below.)
d. 5.83 s.

t0 = 1.00
t = t0 /(1-v2/c2)1/2
t = (1.00 s) /(1-(0.960c)2/c2)1/2
3. Astronaut Rachel leaves Earth in a spaceship at a speed of 0.960c relative to an
observer on Earth. Rachel's destination is a star system 14.4 light-years away (one lightyear is the distance light travels in one year). According to Rachel, how long does the trip
take?
a. 14.4 years.
b 15.0 years.
c. 4.20 years. (See below.)
d. 22.7 years.
L = L0 (1-v2/c2)1/2
L = (14.4 ly) (1-(0.960c)2/c2)1/2 = 4.032 ly
(4.032ly)c/(0.960c) = 4.20 ly
Clayton State University
Department of Natural Sciences
September 8, 2006
Physics 3650 – Quiz 2
Name ______SOLUTION_______________________________
1. Two rocket ships approach Earth from opposite directions, each with a speed of 0.6c
relative to Earth. What is the speed of one ship relative to the other?
V1E = 0.600c
V2E = - 0.600c
V12 = (V1E + VE2)/( 1 + V1E VE2/c2)
V12 = (0.600c + 0.600c))/(1 + (0.600c)(0.600c)/c2) = 0.882c
2. A spaceship and an asteroid are moving in the same direction away from Earth with
speeds of 0.77c and 0.41c, respectively. What is the relative speed between the
spaceship and the asteroid?
VSE = 0.770c
VAE = 0.410c
VSA = (VSE + VEA)/( 1 + VSE VEA/c2)
V12 = (0.770c + (-0.410c)))/(1 + (0.770c)(-0.41 0c)/c2) = 0.526c
Clayton State University
Department of Natural Sciences
September 15, 2006
Physics 3650 – Quiz 3
Name ____SOLUTION_________________________________
4
1. A certain chemical reaction requires 4.82  10 J of energy input for it to go. What is
the increase in mass of the products over the reactants?
E = m c2
m = E / c2
m =(4.82 x 104 J) / (3.00 x 108 m/s)2
m =5.36 x 10-13 kg
2. What is the momentum of a proton traveling at v  0.85c ?
p = m0 v /(1-v2/c2)1/2
m0 = 1.67 x 10-27 kg
p = (1.67 x 10-27 kg)(0.85)( 3.00 x 108 m/s) /(1-(0.85c)2/c2)1/2
p = 8.08 x 10-19 kg-m/s
Clayton State University
Department of Natural Sciences
September 22, 2006
Physics 3650 – Quiz 4
Name __SOLUTION___________________________________
The work function for cesium is 1.90 eV.
a. Find the cutoff frequency and wavelength for the photoelectric effect.
0 = hc/W0
0 = (1240 nm-eV)/(1.90 eV) = 653 nm
f0 = c / 0
f0 = (3.00 x 108 m/s)/( 6.53 x 10-9 m) = 4.59 x 1014 Hz
b. Find the maximum kinetic energy of the electrons if the wavelength of the
incident light is 250 nm.
Kmax = h c/ - W0
Kmax = (1240 nm-eV)/(250 nm) – (1.90 eV) = 3.06 eV = 4.90 x 10-19 J
c. Is the maximum kinetic energy greater or less if the wavelength of the incident
light is 350 nm?
Less, since the photons of 350-nm light carry less energy.
Clayton State University
Department of Natural Sciences
September 29, 2006
Physics 3650 – Quiz 5
Name ___SOLUTION__________________________________
X-rays of wavelength   0.120 nm are scattered from carbon.
a. What is the Compton wavelength shift for photons detected at an angle (relative to
the incident beam) of 45°?
 = h/(mec)(1-cos())
 (2.43 x 10-12 m)(1-cos(45o)) = 0.712 x 10-12 m
b. What is the wavelength of the scattered photon?
 = 0.120 x 10-9 m = 120 x 10-12 m
120 x 10-12 m + 0.712 x 10-12 m = 120.712 x 10-12 m = 0.120712 nm
c. What is the energy of the scattered photon?
E = hc/ = (1240 eV-nm)/(0.120712 nm) = 10,272 eV
Clayton State University
Department of Natural Sciences
October 13, 2006
Physics 3650 – Quiz 6
Name _SOLUTION____________________________________
1. (a) Determine the wavelength of the second Balmer line ( n  4 to n  2 transition).
1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/22 – 1/42) = 0.20625 x 107 m-1
 = 4.85 x 10-7 m = 485 nm
(b) Determine likewise the wavelength of the second Lyman line and
1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/12 – 1/32) = 0.97778 x 107 m-1
 = 1.02 x 10-7 m = 102 nm
(c) the wavelength of the third Balmer line.
1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/22 – 1/52) = 0.231 x 107 m-1
 = 4.33 x 10-7 m = 433 nm
2. What is the speed of an electron in the n = 3 Bohr orbit?
Clayton State University
Department of Natural Sciences
October 30, 2006
Physics 3650 – Quiz 7
Name _SOLUTION____________________________________
Estimate the wavelength for an n  2 to n  1 transition in iron (Z  26).
EK = (Z-1)2(-13.6 eV)/12 = (26-1)2(-13.6 eV)/12 = -8500 eV
EL = (Z-1)2(-13.6 eV)/22 = (26-1)2(-13.6 eV)/22 = -2125 eV
E = 6375 eV
 = hc/E = (1240 eV-nm)/(6375 eV) = 0.195 nm
Clayton State University
Department of Natural Sciences
November 10, 2006
Physics 3650 – Quiz 8
Name ____SOLUTION_________________________________
Estimate the diameter of the following nuclei:
a.
1
1H,
r = (1.20 x 10-15 m) A1/3 = (1.20 x 10-15 m) 11/3 = 1.20 x 10-15 m
d = 2 r = 2.40 x 10-15 m = 2.40 fermi
b.
40
20Ca,
r = (1.20 x 10-15 m) A1/3 = (1.20 x 10-15 m) (40)1/3 = 4.10 x 10-15 m
d = 2 r = 8.20 x 10-15 m = 8.20 fermi
c.
208
82Pb,
r = (1.20 x 10-15 m) A1/3 = (1.20 x 10-15 m) (208)1/3 = 7.11 x 10-15 m
d = 2 r = 14.2 x 10-15 m = 14.2 fermi
d.
235
92U.
r = (1.20 x 10-15 m) A1/3 = (1.20 x 10-15 m) (235)1/3 = 7.41 x 10-15 m
d = 2 r = 14.8 x 10-15 m = 14.8 fermi
Clayton State University
Department of Natural Sciences
November 17, 2006
Physics 3650 – Quiz 9
Name _SOLUTION____________________________________
Find the threshold energy that the incident neutron must have to produce the reaction:
1
0n
+ 42He -->
2
1H
+ 31H
mi = 1.008665 u + 4.002603 u = 5.011268 u
mf = 2.014102 u + 3.016049 u = 5.030151 u
Q = (5.011268 u – 5.030151 u)(931.5 Mev/c2/1 u) = - 17.59 MeV
Kmin = (1 + m/M)|Q| = (1 + (1.008665 u)/(4.002603 u))(17.59 MeV) = 22.023 MeV