Question

What must be the width of a one-dimensional infinite potential well if an electron trapped in it in the $\mathbf{n}\mathbf{=}\mathbf{3}$ state is to have an energy of 4.7 eV ?

Short answer

0.85 nm

Step-by-step solution

Given data

$$\begin{gathered} E_n=4.7eV \\ n=3 \end{gathered}$$

Energy in a potential well

The $n^{th}$ state energy of a particle of mass $m$ in an infinite potential well of width L is

$E_n=\left(\frac{n^2{h}^2}{8m{L}^2}\right)........\left(1\right)$

Determining the width of the potential well

Here h is the Planck's constant having value

$$\begin{gathered} h=6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{and}\mathrm{c}=2.998\times {10}^8\mathrm{m}/\mathrm{s}, \\ {\mathrm{h}}_{\mathrm{c}}=\frac{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(2.998\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\left(1.602\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)\left({10}^{-9}\mathrm{m}/\mathrm{nm}\right)} \\ =1240\mathrm{eV}.\mathrm{nm} \end{gathered}$$

Using the $m{c}^2$value for an electron from Table 37-3 $\left(511\times {10}^3\mathrm{eV}\right)$, Eq.39 – 4 can be rewritten as

$$\begin{gathered} E_n=\frac{n^2{h}^2}{8m{L}^2} \\ =\frac{n^2{\left(h_c\right)}^2}{8\left(m{c}^2\right)L^2} \end{gathered}$$

For $\mathrm{n}=3$,

$$\begin{gathered} L=\frac{n\left(h_c\right)}{\sqrt{8\left(m{c}^2\right)\left(E_n\right)}} \\ =\frac{3\left(1240\mathrm{eV}.\mathrm{nm}\right)}{\sqrt{8\left(511\times {10}^3\mathrm{eV}\right)\left(4.7\mathrm{eV}\right)}} \\ =0.85\mathrm{nm} \end{gathered}$$

Thus, the width is $0.85\mathrm{nm}$.