Physical Chemistry , 1st ed.

quantum number.Because n^2 wavefunctions have the same quantum number

n, the degeneracy of each energy state of the hydrogen atom is n^2. (Again, this

will change by a factor of 2.) Each set of wavefunctions having the same value

for the principal quantum number is said to define a shell.

Example 11.21

Calculate the energy values for the first three shells of the hydrogen atom. The

reduced mass of the hydrogen atom is 9.104 

10 ^31 kg.

Solution

Values are substituted into equation 11.63 for n1, 2, and 3:

E

E

E

These expressions give

E(n1) 2.178 

10 ^18 J

E(n2) 5.445 

10 ^19 J

E(n3) 2.420 

10 ^19 J

where it can easily be shown that the units are joules:



[C^2 /(J

C

m

(^4)
)
k
]
g
 (^2) (J s) 2 

C^4

C

k

4

g

J

2

J

2

s

2

m^2 kg

s

2

m^2 J

Remember that spectroscopy measures the changes in energy between two

states. Quantum mechanics can also be used to determine a change in energy,

E, for the hydrogen atom:

E(n 1 ) E(n 2 ) E

8 

e

(^20)
4
h



 (^2) n 21  8 
e
(^20)
4
h



 (^2) n (^22) 
where the principal quantum numbers n 1 and n 2 are used to differentiate be-
tween the two energy levels involved. A little algebraic rearranging yields

E

8

e



4

(^20)



h^2



n

1

 22 n

1

 (^21)  (11.64)
This is the same form of equation that Balmer got by considering the spectrum
of hydrogen, and that Bohr got by assuming quantized angular momentum! In
fact, the collection of constants multiplying the quantum number expression
is familiar:



8

e



4

(^20)



h^2



2.178

10 ^18 J

(1.602 

10 ^19 C)^4 (9.104 

10 ^31 kg)



8[8.854 

10 ^12 C^2 /(J m]^2 (6.626 

10 ^34 J s)^2

(1.602 

10 ^19 C)^4 (9.104 

10 ^31 kg)



8[8.854 

10 ^12 C^2 /(J m)]^2 (6.626 

10 ^34 J s)^232

(1.602 

10 ^19 C)^4 (9.104 

10 ^31 kg)



8[8.854 

10 ^12 C^2 /(J m)]^2 (6.626 

10 ^34 J s)^222

(1.602^10 ^19 C)^4 (9.104^10 ^31 kg)

8[8.854 

10 ^12 C^2 /(J m)]^2 (6.626 

10 ^34 J s)^212

356 CHAPTER 11 Quantum Mechanics: Model Systems and the Hydrogen Atom