Question

A proton emits a gamma-ray photon with energy 2.0 MeV in a quantum jump from n =2 to n= 1.

Short answer

The length of the box is:

$L=17.5\mathrm{fm}$

Step-by-step solution

Given information

A proton confined in a one-dimensional box emits a 2.0 MeV gamma-ray photon in a quantum jump from n =2 to n= 1.

Explanation

The energy levels of the particle in the box:

$E_n=\frac{h^2{n}^2}{8m{L}^2}$

Energy of the photon:

$\Delta E_{12}=E_2-E_1$

Substitute energy levels:

$$\begin{gathered} E_2-E_1=\frac{h^2·2^2}{8m{L}^2}-\frac{h^2·1^2}{8m{L}^2} \\ {E}_2-E_1=\frac{3h^2}{8m{L}^2} \end{gathered}$$

Calculations:

Express the above equation in terms of L:

$L=\sqrt{\frac{3h^2}{8m\left(E_2-E_1\right)}}$

Substitute the values:

$L=\sqrt{\frac{3{\left(6.63·{10}^{-34}\right)}^2}{8·1.67·{10}^{-27}·2·{10}^6·1.6·{10}^{-19}}}$

$L=17.5\mathrm{fm}$