Department of Natural Sciences
Clayton State University
October 14, 2009
Physics 3650 – Quiz 1
Name _____SOLUTION_______________________________
1. If the displacement of the object, x, is related to velocity, v, according to the
relation x = A v, the constant, A, has the dimension of which of the following?
a. Acceleration.
b. Length.
c. Time.
d. Area.
.
2. If you exert a force F on an object, the force which the object exerts on you will
a. Depend on whether or not the object is moving.
b. Depend on whether or not you are moving.
c. Depend on the relative masses of you and the object.
d. Always be F.
3. What is the SI unit of momentum?
a. N∙m.
b. N/s.
c. N∙s.
d. N/m.
Department of Natural Sciences
Clayton State University
October 19, 2009
Physics 3650 – Quiz 2
Name __SOLUTION___________________________________
1. Astronauts in a spaceship traveling at V = 0.600c relative Earth sign off from
space control saying that they are going to nap for 1 hour and then call back.
a. How long does their nap last as measured on Earth?
t01.00 h

t = t0 /(1-v2/c2)1/2
t = 1.00 s /(1-(0.600c)2/c2)1/2 = 1.25 h
b. Who measures the proper time of the nap?
Astronauts
2. A stick that has a proper length of 1.00 m moves in a direction along its length
with speed V relative to you. The length of the stick as measured by you is 0.914
m. What is the speed V?
L = 0.914 m, L0 = 1.00 m
L = L0 (1-v2/c2)1/2
L/ L0 = (1-v2/c2)1/2
(L/ L0 )2 = (1-v2/c2)
v2/c2 = 1 - (L/ L0 )2
v = 0.406c
Department of Natural Sciences
Clayton State University
October 21, 2009
Physics 3650 – Quiz 3
Name ___SOLUTION__________________________________
1. Spaceship R is moving to the right at a speed of 0.700c with respect to Earth. A
second spaceship, L, moves to the left at the speed of 0.300c with respect to
Earth. What is the speed of L with respect to R?
VRE = 0.700c
VLE = - 0.300c
VLR =?
VLR 
VLE  VER
V V
1  LE 2 ER
c
VSA 
1
 0.300c  0.700c
  0.826c
(0.300c) (0.700c)
c2
Department of Natural Sciences
Clayton State University
October 26, 2009
Physics 3650 – Quiz 4
Name __SOLUTION___________________________________
1. A sodium surface is illuminated with light of wavelength 0.300 m. The work
function for sodium is 2.46 eV. Find
a. The maximum kinetic energy of the ejected photoelectrons.
Kmax hc/ - W0 

Kmax  (1240 eV-nm)/nm - (2.46 eV) = 1.67 eV
b. The cutoff wavelength for sodium.
W0 = hc/0
0 =(1240 eV-nm)/2.46 eV = 504 nm
Department of Natural Sciences
Clayton State University
October 28, 2009
Physics 3650 – Quiz 5
Name ____SOLUTION_________________________________
An X-ray photon of wavelength 5.00 pm that collides with an electron is scattered by an
angle of 60.0o.
a. What is the change in wavelength of the photon?
 = h/(mec)(1-cos())
 (2.43 x 10-12 m)(1-cos(60.0o)) = 0.00122 nm
b. What is the kinetic energy of the scattered electron?
E = hc/ = (1240 eV-nm)/(0.00500 nm) = 248 keV
’0.00500 nm + 0.00122 nm = 0.00622 nm
E’ = hc/’ = (1240 eV-nm)/(0.00622 nm) = 199 keV
K = E – E’ = 49.0 keV
c. What is the magnitude of the scattered photon’s momentum?
p’ = h/’
p’ = (6.63 x 10-34 J-s)/(0.00622 x 10-9 m) = 1.066 x 10-22 kg-m/s
Department of Natural Sciences
Clayton State University
November 6, 2009
Physics 3650 – Quiz 6
Name _SOLUTION____________________________________
A hydrogen atom is in its second excited state according to the Borh model (n = 3).
a. What is the radius of the Bohr orbit?
rn = n2 a0
r5 = (3)2(0.0529 nm) = 0.476 nm
b. What is the angular momentum of the electron?
Ln = n h/(2)
L5 =3 h/(2) = 3.17 x 10-34 J-s
c. What is the electron’s kinetic energy?
L=pr
p = L/r = 6.65 x 10-25 kg-m/s
K = p2/(2me) = (6.65 x 10-25 kg-m/s)2/(2(9.11 x 10-31 kg)) = 2.43 x 10-19 J = 1.52 eV
Alternatively:
E3 = - (13.6 eV)/32 = -1.51 eV
K3 = | E3| = 1.51 eV
d. As the electron “jumps” to the ground state, what is the energy of the
emitted photon?
hf = E3 – E1 = -1.51 eV – (-13.6 eV) = 12.09 eV
e. What is the wavelength of the emitted photon?
hc/ = 12.09 eV
 = 103 nm
Department of Natural Sciences
Clayton State University
November 18, 2009
Physics 3650 – Quiz 7
Name ___SOLUTION__________________________________
1. Estimate the radius of the following nuclei:
a.
7
3Li,
r = (1.20 x 10-15 m) A1/3 = (1.20 x 10-15 m) 71/3 = 2.29 x 10-15 m
b.
209
83Bi.
r = (1.20 x 10-15 m) A1/3 = (1.20 x 10-15 m) 2091/3 = 7.11 x 10-15 m
2. There is a limit to the size of a stable nucleus because of
a. The limited range of the strong nuclear force.
b. The weakness of the electrostatic force.
c. The weakness of the gravitational force.
d. None of the given answers.
Department of Natural Sciences
Clayton State University
November 20, 2009
Physics 3650 – Quiz 8
Name _SOLUTION____________________________________
A certain radioactive nucleus has a half life of 200 s. A sample containing this isotope
has an initial activity of 80,000 Bq.
a. What is the decay constant?
 = ln (2) / T1/2 = ln(2)/(200 s) = 0.003466 1/s
b. How many nuclei were there initially?
Ro =  No
No = Ro/  = (80,000 Bq) /(0.003466 1/s) = 2.31 x 107
c. What is the activity of this sample 10.0 minutes later?
R = Roe- t = (80,000)e-0.003466 1/s (600 s) = 10,000 Bq
Department of Natural Sciences
Clayton State University
November 23, 2009
Physics 3650 – Quiz 9
Name _____SOLUTION________________________________
The mass of
56
Fe is 55.934 9 u and the mass of
56
Co is 55.939 9 u.
.
a. Which isotope decays into the other and by what process? Write the equation for
the decay.
Cobalt-56 decays into Iron-56 by beta decay. (Just compare the masses.)
56
Co26
Fe  e  
56
27
b. How much energy is released in the process?
m = mi – mf = 55.9399 u – 55.9349 u = 0.00500 u
E = m c2 = (0.00500)(931.5 Mev/c2) c2 = 4.66 MeV
Department of Natural Sciences
Clayton State University
December 2, 2009
Physics 3650 – Quiz 10
Name ___SOLUTION__________________________________
Two of the naturally occurring radioactive decay sequences start with
The first five decays of these two sequences are:
,  ,  , , 
and
232
90 Th,
and
235
92 U.
 ,  ,  ,  , .
a. Determine the resulting intermediate daughter nuclei in case of
4
Th228
88 Ra  2 He
232
90

Ra 228
89 Ac  e  
228
88
228
89

Ac 228
90Th  e  
4
Th224
88 Ra  2 He
228
90
4
Ra 220
86 Rn  2 He
224
88
b. work out the
235
92 U.
sequence.
4
U 231
90Th 2 He
235
92

Th231
91Pa  e 
231
90
4
Pa 220
89 Ac  2 He
231
91
220
89

Ac 220
90Th  e  
4
Th216
88 Ra  2 He
220
90
.
232
90 Th,
and
