194
3 Quantum Mechanics – II
Therefore 2αL = 2 × 8.8748 × 10
9
× 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-decay Observed α-energy may be ∼ 5 MeV although the Coulomb
barrier height is 20 or 30 MeV
(ii) Tunnel diode
(iii) Josephson effect In superconductivity electron emission in pairs
through insulator is possible via tunneling mechanism
(iv) Inversion spectral line in ammonia molecule. This arises due to tunneling through the potential barrier between two equilibrium positions of the nitrogen atom along the axis of the pyramid molecule
which is perpendicular to the plane of the hydrogen atoms. The oscillation between the two equilibrium positions causes an intense spectral 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
2m
∂
2
ψ(x, y)
∂ x 2
−
2
2m
∂
2
ψ(x, y)
∂ y 2
= Eψ(x, y)
Let ψ(x, y) = ψ x ψ y
−
2
2m
ψ y
∂
2
ψ x
∂ x 2 −
2
2m
ψ x
∂
2
ψ y
∂ y 2 = Eψ x ψ y
Divide through by ψ x ψ y
−
2
2m
1
ψ x
∂
2
ψ x
∂ x 2 −
2
2m
1
ψ y
∂
2
ψ y
∂ y 2 = E
−
2
2m
1
ψ x
∂
2
ψ x
∂ x 2 − E =
2
2m
1
ψ y
∂
2
ψ y
∂ y 2 = A = constant
∂
2
ψ x
∂ x 2 + α
2
ψ x = 0
where α
2
=
2m
2
(E + A)
ψ x = C sin αx + D cos αx
ψ x = 0 at x = 0
This gives D = 0
ψ x = C sin αx
ψ x = 0 at x = a
This gives αa = n 1 π or α =
n1π
a
Thus ψ x = C sin(n 1 π x/a)
Further
∂
2 ψ y
∂ y 2 =
2mAψ y
2
= −β
2
ψ y
The negative sign on the RHS is necessary, otherwise the ψ y will have an
exponential form which will be unphysical.
ψ y = G sin βy
3 Quantum Mechanics – II
Therefore 2αL = 2 × 8.8748 × 10
9
× 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-decay Observed α-energy may be ∼ 5 MeV although the Coulomb
barrier height is 20 or 30 MeV
(ii) Tunnel diode
(iii) Josephson effect In superconductivity electron emission in pairs
through insulator is possible via tunneling mechanism
(iv) Inversion spectral line in ammonia molecule. This arises due to tunneling through the potential barrier between two equilibrium positions of the nitrogen atom along the axis of the pyramid molecule
which is perpendicular to the plane of the hydrogen atoms. The oscillation between the two equilibrium positions causes an intense spectral 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
2m
∂
2
ψ(x, y)
∂ x 2
−
2
2m
∂
2
ψ(x, y)
∂ y 2
= Eψ(x, y)
Let ψ(x, y) = ψ x ψ y
−
2
2m
ψ y
∂
2
ψ x
∂ x 2 −
2
2m
ψ x
∂
2
ψ y
∂ y 2 = Eψ x ψ y
Divide through by ψ x ψ y
−
2
2m
1
ψ x
∂
2
ψ x
∂ x 2 −
2
2m
1
ψ y
∂
2
ψ y
∂ y 2 = E
−
2
2m
1
ψ x
∂
2
ψ x
∂ x 2 − E =
2
2m
1
ψ y
∂
2
ψ y
∂ y 2 = A = constant
∂
2
ψ x
∂ x 2 + α
2
ψ x = 0
where α
2
=
2m
2
(E + A)
ψ x = C sin αx + D cos αx
ψ x = 0 at x = 0
This gives D = 0
ψ x = C sin αx
ψ x = 0 at x = a
This gives αa = n 1 π or α =
n1π
a
Thus ψ x = C sin(n 1 π x/a)
Further
∂
2 ψ y
∂ y 2 =
2mAψ y
2
= −β
2
ψ y
The negative sign on the RHS is necessary, otherwise the ψ y will have an
exponential form which will be unphysical.
ψ y = G sin βy