1000 Solved Problems in Modern Physics

194 3 Quantum Mechanics – II

Therefore 2αL= 2 × 8. 8748 × 109 × 0. 3 × 10 −^9 = 5. 3249

T= 16

(

2

5

)(

1 −

2

5

)

e−^5.^3249 = 0. 0187

(e) Examples of quantum mechanical tunneling

(i) α-decayObservedα-energy may be∼5 MeV although the Coulomb

barrier height is 20 or 30 MeV

(ii) Tunnel diode

(iii) Josephson effectIn superconductivity electron emission in pairs

through insulator is possible via tunneling mechanism

(iv) Inversion spectral linein ammonia molecule. This arises due to tun-

neling through the potential barrier between two equilibrium posi-

tions of the nitrogen atom along the axis of the pyramid molecule

which is perpendicular to the plane of the hydrogen atoms. The oscil-

lation between the two equilibrium positions causes an intense spec-

tral line in the microwave region.

3.45 The wave function to the zeroeth order in infinitely deep 2-D potential well is
obtained by the method of separation of variables; the Schrodinger equation is

−

^2

2 m

∂^2 ψ(x,y)

∂x^2

−

^2

2 m

∂^2 ψ(x,y)

∂y^2

=Eψ(x,y)

Letψ(x,y)=ψxψy

−

^2

2 m

ψy

∂^2 ψx

∂x^2

−

^2

2 m

ψx

∂^2 ψy

∂y^2

=Eψxψy

Divide through byψxψy

−

^2

2 m

1

ψx

∂^2 ψx

∂x^2

−

^2

2 m

1

ψy

∂^2 ψy

∂y^2

=E

−

^2

2 m

1

ψx

∂^2 ψx

∂x^2

−E=

^2

2 m

1

ψy

∂^2 ψy

∂y^2

=A=constant

∂^2 ψx

∂x^2

+α^2 ψx= 0

whereα^2 =

( 2 m

^2

)

(E+A)

ψx=Csinαx+Dcosαx

ψx= 0 at x= 0

This givesD= 0

ψx=Csinαx

ψx= 0 at x=a

This givesαa=n 1 πorα=n^1 aπ

Thusψx=Csin(n 1 πx/a)

Further∂

(^2) ψy
∂y^2 =
2 mAψy
^2 =−β
(^2) ψy
The negative sign on the RHS is necessary, otherwise theψywill have an
exponential form which will be unphysical.
ψy=Gsinβy