Chapter 6: Q34E (page 226)

Question: Obtain equation (6.18) from(6.16) and (6.17).

Short Answer

Expert verified

Answer

The equation $T = 16 \frac{E_{1}}{U_{0}} (1 - \frac{E_{1}}{U_{0}}) e^{- 2 L \sqrt{2 m (U_{0} - E_{1}) / ℏ}}$ is derived from $α L = \frac{\sqrt{2 m (U_{0} - E)}}{ℏ} L ≫ 1$ and $T = \frac{4 (E / U_{0}) (1 - E / U_{0})}{\sinh^{2} (α L) + 4 (E / U_{0}) (1 - E / U_{0})}$

Step by step solution

Definition of transmission probability

The transmission or tunneling probability can be calculated using transmitted intensity and the incident intensity.

In the case of tunneling barriers being wide, it can be found as follows.

$T ≅ 16 \frac{E}{U_{0}} (1 - \frac{E}{U_{0}}) e^{- 2 L \sqrt{2 m (U_{0} - E_{1}) / ℏ}}$

Here E is the jump energy, U₀is barrier energy, L is the length of the tunnel, and m is the mass of the particle.

Given quantities

The given values are $α L = \frac{\sqrt{2 m (U_{0} - E)}}{ℏ} L ≫ 1$ and $T = \frac{4 (E / U_{0}) (1 - E / U_{0})}{\sinh^{2} (α L) + 4 (E / U_{0}) (1 - E / U_{0})}$ .

Imposing the limiting value in the equation of transmission probability.

We know that,

. $α = \frac{\sqrt{2 m (U_{0} - E)}}{ℏ}$

Use the hyperbolic relation of $Sinh (α L) = \frac{e^{α L}}{2}$ for $α L ≫ 1$ in the transmission formula as:

$T = \frac{4 (E / U_{0}) (1 - E / U_{0})}{\sinh^{2} (α L) + 4 (E / U_{0}) (1 - E / U_{0})} = \frac{4 (E / U_{0}) (1 - E / U_{0})}{\frac{e^{2 α L}}{4} + 4 (E / U_{0}) (1 - E / U_{0})} = \frac{16 (E / U_{0}) (1 - E / U_{0})}{e^{2 α L}} = 16 (E / U_{0}) (1 - E / U_{0}) e^{- 2 α L} = 16 \frac{E}{U_{0}} (1 - \frac{E}{U_{0}}) e^{- 2 L \sqrt{2 m (U_{0} - E) / ℏ}}$