Question

A proton is in a box of width L . What must the width of the box be for the ground level energy to be 5.0 MeV, a typical value for the energy with which the particles in a nucleus are bound? Compare your result to the size of a nucleus-that is, on the order of${\mathbf{10}}^{\mathbf{14}}\mathit{m}$ .

Short answer

The width of the box be$0.641\times {10}^{-14}M$ for the ground level energy to be 5.0 MeV , a typical value for the energy with which the particles in a nucleus are bound and the order of width is same as the diameter of the nucleus.

Step-by-step solution

Define the energy level.

The energy level of a particle of mass in a box of width is:

$$\begin{gathered} {\mathit{E}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{P}}_{\mathbf{n}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{n}}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{n}}^{\mathbf{2}}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}\left(n=1,2,3,...\right) \end{gathered}$$

Determine the width of box.

Given that, energy$E_1$ of photon at ground level is 5.0 MeV .

The energy $E_1$in joule is J

$$\begin{gathered} E_1=5.00\times \left(\frac{1.60\times {10}^{-13}}{1}\right) \\ =8.00\times {10}^{-13}J \end{gathered}$$

The mass of photon is$1.67\times {10}^{-27}kg$ and value of$h=1.055\times {10}^{-34}J.s$ .

The width of the box is:

$$\begin{gathered} L=\frac{n\mathrm{\pi h}}{\sqrt{2m{E}_n}} \\ =\frac{\left(1\right)\mathrm{\pi }\left(1.055\times {10}^{-13}\right)}{\sqrt{2\left(1.67\times {10}^{-27}\right)\left(8.00\times {10}^{-13}\right)}} \\ =0.641\times {10}^{-14}m \end{gathered}$$

Hence, the width of the box is $0.641\times {10}^{-14}m$and order of width is same as the diameter of the nucleus.