Question

The energy of an electron in a $2.00\mathrm{eV}$deep potential well is $1.50\mathrm{eV}$. At what distance into the classically forbidden region has the amplitude of the wave function decreased to $25\%$ of its value at the edge of the potential well?

Short answer

The distance is$3.88\times {10}^{-10}\mathrm{m}.$

Step-by-step solution

Given information 

We have given,

Energy of the electron =$2\mathrm{eV}$

Height of the potential =$1.5\mathrm{eV}$

We have to find the distance where the amplitude will decreases to$25\%$.

Simplify

We know the potential depth is given by,

$$\begin{gathered} H=\frac{h}{2\pi \sqrt{2m(V_0-E)}} \\ H=\frac{6.63\times {10}^{-34}J.s}{2\pi \sqrt{2\times 9.1\times {10}^{-31}\mathrm{kg}(2eV-1.5eV)\times 1.6\times {10}^{-19}\mathrm{J}}} \\ H=2.8{\mathrm{A}}^0 \end{gathered}$$

The forbidden can be found as

$$\begin{gathered} \mathrm{\phi }={\mathrm{\phi }}_{\mathrm{edges}}{\mathrm{e}}^{-(\mathrm{x}-\mathrm{L})/\mathrm{H}} \\ (\mathrm{x}-\mathrm{L})=\mathrm{H}\ln \left(\frac{{\mathrm{\phi }}_{\mathrm{edge}}}{\mathrm{\phi }}\right) \\ (\mathrm{X}-\mathrm{L})=(2.8\times {10}^{-10}\mathrm{m})\ln \left(\frac{1}{0.25}\right) \\ (\mathrm{x}-\mathrm{L})=(2.8\times {10}^{-10}\mathrm{m})(1.386) \\ (\mathrm{x}-\mathrm{L})=3.88{\mathrm{A}}^0 \end{gathered}$$