Question

A typical electron in a piece of metallic sodium has energy$-{\mathrm{E}}_0$compared to a free electron, where ${\mathrm{E}}_0$is the$2.7\mathrm{eV}$ work function of sodium.

a. At what distance beyond the surface of the metal is the electron’s probability density $10\%$of its value at the surface?

b. How does this distance compare to the size of an atom?

Short answer

a.$1.4{\mathrm{A}}^0$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

${\mathrm{E}}_0=2.7\mathrm{eV}$

$\mathrm{P}=10\%$

we have to find the distance.

Simplify

We know that,

${\mathrm{P}}_{\mathrm{tunnel}}={\mathrm{e}}^{-2\mathrm{w}/\mathrm{\eta }}$

and,

$$\begin{gathered} \mathrm{\eta }=\frac{\mathrm{h}}{2\mathrm{\pi }\sqrt{2\mathrm{m}({\mathrm{U}}_0-\mathrm{E})}} \\ \mathrm{\eta }=\frac{(6.62\times {10}^{-34}\mathrm{J}.\mathrm{s})}{2\mathrm{\pi }\sqrt{2\times 9.1\times {10}^{-31}\mathrm{kg}(2.7\times 1.6\times {10}^{-19}\mathrm{J})}} \\ \mathrm{\eta }=0.12\mathrm{nm} \end{gathered}$$

Put this value in probability equation

$$\begin{gathered} \mathrm{P}={\mathrm{e}}^{-2\mathrm{w}/(0.12)}=0.1 \\ \frac{-2\mathrm{w}}{0.12}=-2.3 \\ \mathrm{w}=0.14\mathrm{nm} \\ \mathrm{w}=1.4{\mathrm{A}}^0 \end{gathered}$$

Part (b) Step 1: Given information

We have to find its distance compare with the atom size.

Simplify 

Since the found distance is comparable to the size of the atom because the most atoms have to size in${\mathrm{A}}^0.$