Question

A proton is confined to a one-dimensional infinite potential well 100pm wide. What is its ground-state energy?

Short answer

0.0205eV

Step-by-step solution

Given data

The width of the potential well is

$$\begin{gathered} L=100pm \\ =100\times \left(1pm\times \frac{1m}{{10}^{12}pm}\right) \\ =100\times {10}^{-12}m \end{gathered}$$

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 proton

The mass of the proton is

$m_p=1.67\times {10}^{-27}kg$

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

$$\begin{gathered} E_1=\frac{\left(6.63\times {10}^{-34}J.s\right)}{8\times 1.67\times {10}^{-27}kg\times \left(100\times {10}^{-12}m\right)}{\left(1\right)}^2 \\ =3.29\times {10}^{-21}\times \left(1J\right).\left(1J\times \frac{1kg.m^2/s^2}{1J}\right).\left(1s^2\right).\left(\frac{1}{1kg}\right).\left(\frac{1}{m^2}\right) \\ =3.29\times {10}^{-21}J \end{gathered}$$

The energy in electron volt is

$$\begin{gathered} E_1=3.29\times {10}^{-21}J\times \left(1J\times \frac{0.624\times {10}^{19}eV}{1J}\right) \\ =0.0205eV \end{gathered}$$

The required energy is 0.0205 eV .