CHEM106: Assignment 8
Hydrogen Atom
1. For the hydrogen-like atoms, the solution of the Schrödinger equation leads to the
m Z 2e4
quantized energy En   e 2 2 2 .
8 0 h n
A. Calculate the energy (in eV) for the hydrogen atom at various values of n, and
plot the energy levels.
For the hydrogen atom, z = 1, and E n  
me e 4
R
  H2 , and the Rydberg
2 2 2
n
8 0 h n
constant RH is defined as
m e4
9.10938  10 31 kg  (1.60218  10 19 C ) 4
RH  e2 2 
 2.1788  10 18 J
12 2 1 1 2
 34
2
8
(
8
.
85418

10
C
J
m
)
(
6
.
62607

10
Js
)
8 0 h
Given that 1eV =1.60218´10-19 J = 8065.73cm-1 , we have
RH = 2.1788´10-18 J =13.59eV =10973.73cm-1
mee4
R
13.6
Thus, En = - 2 2 2 = - H2 = - 2 (eV ) .
8e0 h n
n
n
n
En(eV)
1
–13.59
2
–3.40
3
–1.51
4
–0.85
5
–0.54
∞
0
The energy levels can be plotted as follows
B. Calculate the absorption frequency in wavenumbers (cm–1) for an electron in the
hydrogen atom undergoing a transition from the n = 2 level to the n = 5 level.
Look up the energy table in part A. We have
DE = E5 - E2 = -0.54eV - (-3.40eV) = 2.86eV = 4.58´10-19 J .
Since E photon  h  E ,
E
4.58  10 19 J

 6.912  1014 s 1 .
 34
h
6.62607  10 Js
To express the frequency in wavenumber, we need to divide the
frequency by the speed of light
n
6.912 ´1014 s-1
n= =
= 23060cm-1
10
-1
c 2.99793´10 cm × s
we have  
C. Determine the ionization energy (in eV) for the hydrogen atom.
As the hydrogen atom ionizes, the electron moves from the n = 1 state (ground
state) to n = ∞ state. Look up the energy table in part A. We have
IE = E¥ - E1 = 0 - (-13.59eV) =13.59eV .
2. The solution of the Schrödinger equation for the hydrogen atom produces three
quantum numbers n, l, and ml.
A. Name the three quantum numbers, and describe how they are related.
The three quantum numbers n, l, and ml are named principal quantum number,
angular momentum quantum, and magnetic quantum number, respectively.
For a given n, l  n  1 .
And for a given l, ml = -l,-(l -1),...,-1, 0,1,...,(l -1),l .
B. For the n = 3 energy level of the hydrogen atom, what are the allowed quantum
states?
The allowed quantum states are characterized by the three quantum numbers n,
l, ml.
For n = 3, the possible combinations (n, l, ml) are (3, 2, –2), (3, 2, –1),
(3, 2, 0), (3, 2, 1), (3, 2, 2), (3, 1, –1), (3, 1, 0), (3, 1, 1), and (3, 0, 0).
C. For the n = 3 energy level of the hydrogen atom, what is the degeneracy?
For the hydrogen atom, energy is only dependent on the principal quantum
number. For the n = 3 energy level, as shown in part B above, there are nine
possible states. Thus, the degeneracy is 9. In general, the degeneracy for the
nth energy level is n2.
3. The solution of the Schrödinger equation for the hydrogen atom results in the
r
1 3 
wavefunction for the 1s state 1s  100 
( ) 2 e a0 .
 a0
1
A. Calculate the probability of finding the electron within a distance of 3a0 from the
nucleus.
The probability for finding the electron is given by
2

0
0
p(0  r  3a0 )   d 
sin d
3a 0

2

0
0
1s 1s r dr   d 
*
2
0
sin d
3a 0

0
2r
1  a0 2
e r dr
a0 3
2r

a0
3a 0
1
e r 2 dr
3
0 a0
Let z = r / a0, we have
4
3
p  4 e  2 z z 2 dz
0
  e  2 z z 2 dz  
1 2 2 z
1
1
z de   ( z 2 e  2 z  2  e  2 z z dz )   ( z 2 e  2 z   z de  2 z )

2
2
2
1
1
1
e 2 z 2
1
  ( z 2 e  2 z  ze  2 z   e  2 z dz )   ( z 2 e  2 z  ze  2 z  e  2 z )  
(z  z  )
2
2
2
2
2
Here, a mathematical technique of integration by parts,
 udv  uv   vdu
is
adopted.
3
 p  4 e
0
3
2 z
e 2 z 2
1
z dz  4  [ 
( z  z  )]  1  e  6 ( 2  32  2  3  1)  0.94  94%
2
2 0
2
B. Is there a node in the wavefunction for the 1s orbital?
A node is a point, plane or surface at which the wavefunction is zero, and
changes sign before and after. The only possibility for the wave function
r
1 1 23  a0
to be zero is when r approaches infinite. Thus, there is no
1s 
( ) e
 a0
node in the 1s orbital.
4. What is the meaning of the atomic term symbol 3F?
The electronic states of a multi-electron atom can be characterized by the term
symbol 2S+1L, where L and S are total angular momentum and total spin quantum
number, respectively. The term symbol 3F means that L = 3 and S = 1.