Review Exam #2
1. A monochromatic beam of light is absorbed by a collection of groundstate hydrogen atoms in such a way that six different wavelengths are
observed when the hydrogen relaxes back to the ground state.
(a) What is the wavelength of the incident beam?
(b) What is the longest wavelength in the emission spectrum of these
atoms?
(c) To what portion of the electromagnetic spectrum does it belong?
(d) To what series does it belong?
(e) What is the shortest wavelength?
(f) To what series does it belong?
Solution:
(a) The batch of excited atoms must make these six transitions to get back
to state one: 2  1 , and also 3  2 and 3  1 , and also 4  3 and
4  2 and 4  1 . Thus, the incoming light must have just enough energy
to produce the 1  4 transition. It must be the third line of the Lyman
series in the absorption spectrum of hydrogen. The absorbing atom
changes from energy
Ei  
13.6 eV
13.6 eV


13.6
eV
E


 0.850 eV
f
to
12
42
so the incoming photons have wavelength  = c/f = hc/Ephoton
6.626  1034 J  s  3.00  108 m s   1.00 eV 

hc



.
19
E f  Ei
0.850 eV   13.6 eV 
 1.60  10 J 
 9.74  108 m  97.4 nm
(b) The longest of the six wavelengths corresponds to the lowest photon
13.6 eV
 0.850 eV
energy, emitted in the 43 transition. Here Ei  
42
and Ef  
13.6 eV
 1.51 eV ,
32
so

hc
1240 eV  nm

 1.88  m
Ef  Ei 0.850 eV   1.51 eV 
.
This infrared wavelength is part of the Paschen series, since the lower
state has n = 3.
(c) The shortest wavelength emitted is the same as the wavelength
absorbed: 97.4 nm, ultraviolet, Lyman series.
2. (a) How much energy is required to cause an electron in hydrogen to
move from the n = 1 state to the n = 2 state?
(b) Suppose the electron gains this energy through collisions among
hydrogen atoms at a high temperature. At what temperature would the
average kinetic energy 3kBT/2, where kB is the Boltzmann constant, be
great enough to excite the electron?
Solution:
(a)
The energy difference between these two states is equal to the
energy that is absorbed.
Thus,
E  E2  E1 
(b)
3
E  kBT
2
 13.6 eV   13.6 eV 
4

1
 10.2 eV  1.63  1018 J
2 1.63  1018 J
2E
T

 7.88  104 K
23
3kB 3 1.38  10 J K 
3. Suppose the ionization energy of an atom is 4.10eV. In the spectrum of
this same atom, we observe emission lines with wavelengths 310nm, 400
nm, and 1377.8 nm. Use this information to construct the energy level
diagram with the fewest levels. Assume that the higher levels are closer
together. Sketch this energy level diagram.
Solution:
E
hc


1 240 eV  nm
1  310 nm ,
2  400 nm ,
3  1 378 nm ,

 E
so E1  4.00 eV
E2  3.10 eV
E3  0.900 eV
and the ionization energy  4.10 eV .
The energy level diagram having the fewest levels and consistent
with these energies is shown below
4. The positron is the antiparticle to the electron. It has the same mass and
a positive electric charge of the same magnitude as that of the electron.
Positronium is a hydrogen-like atom consisting of a positron and an
electron revolving around each other. Using the Bohr model, find the
allowed distances between the two particles and the allowed energies of
the system.
Solution:
Let r represent the distance between the electron and the positron. The two
r
move in a circle of radius
around their center of mass with opposite
2
velocities. The total angular momentum of the electron-positron system is
quantized to according to
Ln 
n  1,2,3,
where
For each particle,
m vr m vr

n
2
2
.
kee2
 F  m a expands to
n
v

We can eliminate
m r to find
r2
m v2

r2 .
kee2 2m n2 2

r
m 2r2 .
So the separation distances are
2n2 2
2
r

2
a
n

0
m kee2
1.06 10
10

m n2 .
r
2
The orbital radii are 2  a0n , the same as for the electron in hydrogen.
The energy can be calculated from
1 2 1 2 kee2
E  K U  mv  mv 
2
2
r
Since
kee2
mv 
,
2r
2
.
kee2 kee2
kee2 kee2
6.80 eV
E





.
2r
r
2r 4a0n2
n2