1000 Solved Problems in Modern Physics

162 3 Quantum Mechanics – II

3.12 First the wave function is normalized

N^2

∫∞

0

ψ∗ψdx= 1

N^2

∫∞

0

(√

2 e−

x

L

) 2

dx= 1

N= 1 /

√

L

The probability of finding the particle in the regionx≥1nmis

(

1

L

)∫∞

1

ψ∗ψdx=

∫∞

1

((

1

L

)^12

e−

xL

) 2

dx=

(

2

L

)∫∞

1

e−^2 x/Ldx

=−e−^2 x/L

∣

∣∞

1 =e

− (^2) = 0. 135

3.3.2 Schrodinger Equation. .........................

3.13

(

d^2

dr^2

+

2

r

+ 2 E

)

F(r)=0(1)

(a) By usingF(r)=exp(−r/v)y(r), andE=− 21 ν 2 , it is easily verified that

d^2 y

dr^2

=

2

v

(

d

dr

−

v

r

)

y (2)

(b)y(r)=

∑∞

p= 0

aprp+^1 (3)

dy

dr

=

∑

ap(p+1)rp (4)

d^2 y

dr^2

=

∑

app(p+1)rp−^1 (5)

Substitute (3), (4) and (5) in (2)

Σapp(p+ 1 )rp−^1 =

2

v

Σap(p+1)rp− 2 Σaprp

Replacepbyp−1 in the RHS and simplify

Σapp(p+1)rp−^1 =

2

v

Σap− 1 (p−ν)rp−^1

Comparing the coefficients ofrp−^1 on both sides

p(p+1)ap=

2

v

(p−v)ap− 1 (6)

(c) The series in (3) will terminate whenν=nwherenis a positive integer.

Heren= 2

Using (3)

y(r)=

∑^1

0

ap rp+^1 =a 0 r+a 1 r^2