1000 Solved Problems in Modern Physics

210 3 Quantum Mechanics – II

∇^2 =

d^2

dr^2

+

(

2

r

)

d

dr

(2)

Inserting (2) in (1) and performing the integration we get

<T>=

^2

2 ma 02

Buta 0 = 

2

me^2

Or

2

m=a^0 e

2

Therefore,<T>= e

2

2 a 0

Also<E>=<T>+<V>=e

2

2 a 0 −

e^2

a 0 =−e

(^2) / 2 a 0
3.65 The normalized eigen function for the ground state of hydrogen atom is
ψ 0 = 1 /

(

πa^30

)^12

e−r/a^0

wherea 0 is the Bohr’s radius

(a) The probability that the electron will be formed in the volume element dτ

is

p(r)dr=|ψ 0 |^2 dτ=

(

e

−a^2 r 0

πa^30

)

. 4 πr^2 dr

=

(

4

a 03

)

r^2 e−^2 r/a^0 dr

Maximum probability is found by settingddpr= 0

d

dr

(

4 r^2 e−

(^2) ar
0
a 03

)

=

(

8 r

a^30

)

e−

(^2) ar
0

(

1 −

r

a 0

)

= 0

Thereforer=a 0

(b)<r>=

∫∞

0

ψ∗rψdτ=

1

π

a^30

∫∞

0

rexp

(

−

2 r

a 0

)

. 4 πr^2 dr

=

(

4

a 03

)∫∞

0

r^3 exp

(

−

2 r

a 0

)

dr

Let

2 r

a 0

=x;dr=

a 0 dx

2

<r>=

(a

0

4

)∫∞

0

x^3 e−xdx=

(a

0

4

)

3!=

3 a 0

2

3.66

∫

u^2210 dτ=A 22

∫

e−^2 xx^2 cos^2 θr^2 sinθdθdrdφ

=

(

1

π

)(

1

2 a 0

) 3 ∫∞

0

exp

(

−

r

a 0

)(

r^2

4 a 02

)

r^2 dr

∫+ 1

− 1

cos^2 θd(cosθ)

∫ 2 π

0

dφ

where we have putx=r/ 2 a 0. Putr/a 0 =y,dr=a 0 dy