PHY 3101
Spring 2008
Exam #2
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1. An isolated atom of a certain element emits light of wavelength 520 nm
when the atom falls from its fifth excited state into its second excited state.
The atom emits a photon of wavelength 410 nm when it drops from its sixth
excited state into its second excited state. Find the wavelength of the light
radiated when the atom makes a transition from its sixth to its fifth excited
state.
Solution:
The fifth excited state must lie above the second excited state by the photon
energy
6.63  1034 J s 3  108 m s
E52  hf 

 3.82  1019 J
9

520  10 m
hc
E52 
hc


1240eV  nm
 2.38eV
520nm
The sixth excited state exceeds the second in energy by
6.63  1034 J s 3  108 m s
19
E62 

4.85

10
J
410  109 m
hc 1240eV  nm
E62 

 3.02eV

410nm
Then the sixth excited state is above the fifth by
 4.85  3.82 1019 J  1.03  1019 J or (3.02 – 2.38) eV = 0.64 eV
In the 6 to 5 transition the atom emits a photon with the infrared wavelength
hc (6.63 1034 J  s)(3 108 m / s)
6



1
.
94

10
m
19
E65
1.03 10 J

hc 1240eV  nm

 1937nm
E65
0.64eV
2. A hydrogen atom is in its third excited state (n = 4). Using the Bohr’s theory
of the atom, calculate the following:
(a) the radius of the orbit
(b) the linear momentum of the electron
(c) the angular momentum of the electron
(d) the kinetic energy of the electron
(e) the potential energy of the system
(f) the total energy of the system
Solution:
r42
(a)
(b)
mev4 
 a0n2  (0529nm)(42 )  0.846nm
m kee2
r4

(9.11 1031kg)(8.99  109 N  m2 / C 2 )(1.6  1019 C )2
mev4  4.97  1025
(c)
0.846  109 m
kg  m
s
2

 25 kg  m 
9
34 kg  m
L4  mev4r4   4.97  10
(0.846  10 m)  4.20  10
s
s


2

 25 kg  m 
4
.
97

10


1
(mev4 )2 
s 
2
19
K 4  mev4 


1
.
35

10
J  0.84eV
(d)
2
2me
2(9.11  1031kg)
(e)
ke e 2
(8.99 109 N  m 2 / C 2 )(1.6 10 19 C ) 2
U4  

 2.72 10 19 J  1.70eV

9
r4
0.846 10 m
(f)
E4  K4  U4  0.84eV  1.70eV  0.84eV
3. A photon with energy 2.28 eV is barely capable of causing a photoelectric
effect when it strikes a sodium plate. Suppose the photon is instead absorbed
by hydrogen.
(a) Find the minimum n for a hydrogen atom that can be ionized by such a
photon.
(b) Find the speed of the released electron far from the nucleus.
Solution:
(a)
The photon has energy 2.28 eV.
13.6 eV
 3.40 eV is required to ionize a hydrogen atom from
And
22
state n  2 . So while the photon cannot ionize a hydrogen atom pre-excited
to n  2 , it can ionize a hydrogen atom in the n  3 state, with energy

13.6 eV
 1.51 eV .
32
(b)
The electron thus freed can have kinetic
1
2
energy Ke  2.28 eV  1.51 eV  0.769 eV  mev . Therefore
2
v
2  0.769 1.60  1019  J
9.11 1031 kg
 520 km s .
4. A .00160-nm photon scatters from a free electron. For what (photon)
scattering angle does the recoiling electron have kinetic energy equal to the
energy of the scattered photon?
Solution:
K e  E0  E gives
E  E0  E
hc
E
hc
hc
 
 2  20
E  0 and   
E0 2
E0
E
2
   0   C  1 cos  20  0  C  1 cos 
With K e  E  ,
1 cos 
0 0.00160

C 0.002 43
  70.0
Formulas:
1. The first Bohr’s radius (for electron in hydrogen atom in the ground state):
a0 
2
2
 0.0529nm
mke
2. The ground energy level in hydrogen atom:
E0 
mk 2e 4
2 2
1 ke2

 13.6eV
2 a0
3. The energy levels, orbit radius, kinetic energy, and momentum of electron
in an atom:
2
2
1
kZe

a
2
2 E0
2
2 0
K E  mv 
En   Z 2
r

n

n
n
2
2r
2
Z
mkZe
p  me v 
Ze2
U  k
r
me ke e 2
r
4. Angular momentum:
L  mvr
5. The Bragg condition:
2d sin   n ,
n  1,2,3,4.....
6. The Compton Effect:
C 
h
 
1 cos 
m ec
h
me c
E 
7. The average energy of a particle in a gas:
8. Useful Formulas for Wavelength and Energy:
E  hf 
hc

f 
c

3
k BT
2
9. Photoelectric Effect:
 mv2 

eV0  
 hf  
 2 

 max
10. Plank’s Law:
u ( ) 
c 
hc

8hc 25
hc
e kT
11. Stefan-Boltzman Radiation Law:
1
R
P
 T 4
A
12. Wien’s displacement law:
max T  cons tan t  2.898 103 m  K
kB = 1.38 x 10-23J/K
hc = 1240 eV/nm
ћ = 1.055x10-34J·s
NA= 6.02 x 1023 mol-1
1J = 6.24x1018eV;
c = 3.00x108 m/s
ke = 8.98x109 Nm2/C2
melectron =9.11 x10-31kg
e = -1.6 x 10-19C
h = 6.62 x 10-34J.s=4.136 x 10-15eV∙s
1eV = 1.602x10-19J
mneutron = 1.67x10-27kg
kB= 1.38 x 10-23J/K