Answer
Let for electron has wave function
ψ ∝ e i k x ℏ \psi\propto e^\frac{ikx}{\hbar} ψ ∝ e ℏ ik x
The effective momentum of electron is given by
ℏ 2 k 2 2 m ∗ = − q ϕ \frac{\hbar^2 k^2}{2m*}=-q\phi 2 m ∗ ℏ 2 k 2 = − qϕ
k = i 2 m ∗ q ϕ ℏ 2 k=i\sqrt{\frac{2m*q\phi}{\hbar^2}} k = i ℏ 2 2 m ∗ qϕ
So wavefunction become
ψ ∝ e i 2 m ∗ q ϕ ℏ 2 x \psi\propto e^{i\sqrt{\frac{2m*q\phi}{\hbar^2}}x} ψ ∝ e i ℏ 2 2 m ∗ qϕ x
At x=0 to x=d the ratio of wave functions is given
ψ ( x = d ) ψ ( x = 0 ) ∝ e − 2 m ∗ q ϕ ℏ 2 . d e − 2 m ∗ q ϕ ℏ 2 . 0 ∝ e − 2 m ∗ q ϕ ℏ 2 d \frac{\psi (x=d) }{\psi(x=0) }\propto \frac{ e^{-\sqrt{\frac{2m*q\phi}{\hbar^2}}.d}}{e^{-\sqrt{\frac{2m*q\phi}{\hbar^2}}.0}}\propto
e^{-\sqrt{\frac{2m*q\phi}{\hbar^2}}d} ψ ( x = 0 ) ψ ( x = d ) ∝ e − ℏ 2 2 m ∗ qϕ .0 e − ℏ 2 2 m ∗ qϕ . d ∝ e − ℏ 2 2 m ∗ qϕ d
Let JI
be the incident current and JT
be the transmitted current. Then
Transmission probability
T ∝ J T J 0 ∝ ( ψ ( x = d ) ψ ( x = 0 ) ) 2 ∝ e − 2 − 2 m ∗ q ϕ ℏ 2 d T\propto \frac{J_T}{J_0}\propto (\frac{\psi(x=d) }{\psi(x=0) }) ^2\\\propto e^{-2\sqrt{-\frac{2m*q\phi}{\hbar^2}}d} T ∝ J 0 J T ∝ ( ψ ( x = 0 ) ψ ( x = d ) ) 2 ∝ e − 2 − ℏ 2 2 m ∗ qϕ d
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