PHYS 212-071
HW # 5 Solutions
(Numbers refer to 2nd Edition)
18. A hydrogen atom initially in its ground state ( n  1) absorbs a photon and ends up in
the state for which n  3 . (a) What is the energy of the absorbed photon? (b) If the atom
returns to its ground state, what photon energies could the atom emit?
(a) E1  13.6eV , E3  
13.6eV
 1.5eV  E Photon  E3  E1  12.1eV
32
(b) The photon returns to the ground state either directly by emitting one photon of
energy 12.1 eV, or return first to the n = 2 state by emitting a photon of energy
E3  E2  1.5  (3.4)  1.9eV and then from the n = 2 state to the ground state by
emitting a photon of energy E2  E1  3.4  (13.6)  10.2eV
23. A hydrogen atom is in its ground state ( n  1). Using the Bohr theory of the atom,
calculate (a) the radius of the orbit, (b) the linear momentum of the electron, (c) the
angular momentum of the electron, (d) the kinetic energy, (e) the potential energy, and (f)
the total energy.

(a) rn  n 2 a0  r1  a0  0.529 A
(b) pc   2mc 2 E  2  0.511  10 6  13.6  3728eV  p  2  10 26 kg.m / s
(c) L    6.582  10 16 eV .s
(d) KE  E  13.6eV
(e) PE  27.2eV
(d) E  13.6eV
26. Calculate the longest and shortest wavelengths in the Lyman series for hydrogen,
indicating the underlying electronic transition that gives rise to each. Are any of the
Lyman spectral lines in the visible spectrum? Explain.
1 
1
 R 2  2 , n  2,3,4,..., R  1.097  10 7 m 1

n 
1
1
1
 Longest 
4
1
 121nm ,  shortest   91nm . None of the Lyman lines is in the visible
3R
R
spectrum.
37. Use Bohr’s model of the hydrogen atom to show that when an atom makes a
transition from the state n to the state n  1 , the frequency of the emitted light is given by
f 
2 2 me k 2 e 4  2n  1 

2 2
h3
 n  1 n 
Show that as n   , the expression above varies as 1
n3
and reduces to the classical
frequency one would expect the atom to emit. (Hint: To calculate the classical frequency,
note that the frequency of revolution is
v
n2 2
, where r is given by rn 
). This is an
2r
me ke2
example of the correspondence principle which requires that the classical and quantum
models agree for large values of n .
mvr  n
ke2 mv 2
n2 2

 rn 
r
r2
mke2
ke2
ke2 mke2
k 2 me 1
1 2
ke2
En  

 2 2 

  KE   mv  v 
2rn
2
2
n
n 
2 2 n 2
ke2
v
mk 2 e 4
4 2 mk 2 e 4 1
n
f 



 3
3 3
3
2r 2  n 2  2
2


n
h
n
mke2
We also have
E  hf  En  En1 
k 2 me 4
2 2
 1
1 
2 2 mk 2 e 4




f

 n  12 n 2 
h3


2
 2n  1 
4 2 mk 2 e 4 1



 3
3
 n  12 n 2  n
h
n

