 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-2/section-b/21/ii/c/solution

For a finite step and equal masses, continuity of the wavefunction and its derivative gives

A + B = C, k_{1} (A - B) = k_{2} C .

Hence the reflected and transmitted amplitude ratios are

r = \frac{B}{A} = \frac{k _{1} - k _{2}}{k _{1} + k _{2}}, t = \frac{C}{A} = \frac{2 k _{1}}{k _{1} + k _{2}} .

j = (ℏ k / m) ∣ A ∣^{2}

R = ∣ r ∣^{2} = (\frac{k _{1} - k _{2}}{k _{1} + k _{2}})^{2}, T = \frac{k _{2}}{k _{1}} ∣ t ∣^{2} = \frac{4 k _{1} k _{2}}{( k _{1} + k _{2} ) ^{2}} .

They satisfy

R + T = 1

. Using

∣ t ∣^{2}

alone as the transmission probability would omit the different transmitted velocity.

For

E = 2 eV

and a downward step of

2 eV

, the transmitted kinetic energy is

4 eV

and

k_{2} / k_{1} = 2

. Therefore

R = 0.0294 (2.94%), T = 0.9706 (97.1%) .

The nonzero reflection is compatible with current conservation even though the potential drops.

 New to topics? Read the docs here!