Question

Consider an atomic nucleus to be equivalent to a one dimensional infinite potential well with $\mathit{L}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}$, a typical nuclear diameter. What would be the ground-state energy of an electron if it were trapped in such a potential well? (Note: Nuclei do not contain electrons.)

Short answer

1.9 GeV

Step-by-step solution

Given data

The width of the potential well is

$L=1.4\times {10}^{-14}m$

Energy in a potential well

The ground state energy of a particle of mass m in an infinite potential well of width L is

$E_0=\left(\frac{h^2}{8m_e{L}^2}\right)n^2............\left(1\right)$

Here h is the Planck's constant having value

$h=6.63\times {10}^{-34}J.s$

Determining the ground state energy of electron

The mass of the electron is

$m_e=9.11\times {10}^{-31}kg$

From equation (I) the ground state energy of electron is

$$\begin{gathered} E_1=\frac{{\left(6.63\times {10}^{-34}J.s\right)}^2}{8\times 9.11\times {10}^{-31}kg\times {\left(1.4\times {10}^{-14}m\right)}^2}{\left(1\right)}^2 \\ =3.07\times {10}^{-10}\times \left(1J\right).\left(1J\times \frac{1kg.m^2/s^2}{1J}\right).\left(1s^2\right).\left(\frac{1}{1kg}\right).\left(\frac{1}{1m^2}\right) \\ =3.07\times {10}^{-10}J \end{gathered}$$

The energy in electron volt is

$$\begin{gathered} E_1=3.07\times {10}^{-10}\times \left(1J\times \frac{0.624\times {10}^{19}eV}{1J}\right) \\ =1.9\times {10}^{9}eV \\ =1.9GeV \end{gathered}$$

The required energy is $1.9GeV$.