bei48482_FM

Since dV r^2 sindr ddwe have for the expectation value of 1 r

^



0 

^2 dV

The integrals have the respective values



0

re^2 r^ a^0 dr e^2 r^ a^0  e^2 r^ a^0 

0







0

sin d[cos ] 0  2

2 

0

d[] 02  2 

Hence (2)(2)

Example 6.3

How much more likely is a 1selectron in a hydrogen atom to be at the distance a 0 from the

nucleus than at the distance a 0     2?

Solution

According to Table 6.1 the radial wave function for a 1selectron is

R er^ a^0

From Eq. (6.25) we have for the ratio of the probabilities that an electron in a hydrogen atom

be at the distances r 1 and r 2 from the nucleus



Here r 1 a 0 and r 2 a 0   2, so

 4 e^1 1.47

The electron is 47 percent more likely to be a 0 from the nucleus than half that distance (see

Fig. 6.11).

Angular Variation of Probability Density

The function varies with zenith angle for all quantum numbers land mlexcept

lml0, which are sstates. The value of  ^2 for an sstate is a constant; ^12 , in fact.

This means that since  ^2 is also a constant, the electron probability density ^2 is

(a 0 )^2 e^2



(a 0     2)^2 e^1

Pa 0



Pa 0    2

r^21 e^2 r^1 a^0



r^22 e^2 r^2 a^0

r^21 R 1 ^2



r^22 R 2 ^2

P 1



P 2

2



a 03    2

1



a 0

a^20



4

1



a^30

1



r

a^20

4

r



2

a^20



4

1



r

1



r

Quantum Theory of the Hydrogen Atom 215





0

re^2 r^ a^0 dr



0

sind

2 

0

d

1



a^30

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