Question

A particle confined in a rigid one-dimensional box of length $10\mathrm{fm}$has an energy level ${\mathrm{E}}_{\mathrm{n}}=32.9\mathrm{MeV}$and an adjacent energy level ${\mathrm{E}}_{\mathrm{n}+1}=51.4\mathrm{MeV}$.

a. Determine the values of n and n + 1.

b. Draw an energy-level diagram showing all energy levels from 1 through n + 1. Label each level and write the energy beside it.

c. Sketch the n + 1 wave function on the n + 1 energy level.

d. What is the wavelength of a photon emitted in the $\mathrm{n}+1\to \mathrm{n}$transition? Compare this to a typical visible-light wavelength.

e. What is the mass of the particle? Can you identify it?

Short answer

here,$\mathrm{n}=4$and the energy diagram is given below.

Wavelength is not in visible range and the particle is seem to be proton.

Step-by-step solution

Part (a) step 1: Given information

We have given,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}=32.9\mathrm{MeV} \\ {\mathrm{E}}_{\mathrm{n}+1}=51.4\mathrm{MeV} \end{gathered}$$

we have to find the value of n and n+1.

Simplify

Since the energy is one dimensional potential box is given by,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{h}}^2{\mathrm{n}}^2}{8{\mathrm{mL}}^2} \\ {\mathrm{E}}_{\mathrm{n}+1}=\frac{{\mathrm{h}}^2{(\mathrm{n}+1)}^2}{8{\mathrm{mL}}^2} \end{gathered}$$

Divided the both equation

$$\begin{gathered} \frac{{\mathrm{E}}_{\mathrm{n}+1}}{{\mathrm{E}}_{\mathrm{n}}}=\frac{{(n+1)}^2}{n^2} \\ \frac{51.4}{32.9}=\frac{{(n+1)}^2}{n^2} \\ 1.25n=n+1 \\ 0.25n=1 \\ n=4 \\ n+1=5 \end{gathered}$$

Part (b) step 1: Given information

We have to draw the energy level diagram.

Simplify

Part (c) Step 1: Given information

We have to sketch the wave functions.

Simplify

Part (d) Step 1: Given information

We have to find the wavelength of the wave.

Simplify

We know that we can write,

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{n}\to \mathrm{m}}=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{n}}-{\mathrm{E}}_{\mathrm{m}}} \\ {\mathrm{\lambda }}_{5\to 4}=\frac{1240\mathrm{eV}.\mathrm{nm}}{(51.4-32.9)\times {10}^6\mathrm{eV}} \\ \mathrm{\lambda }=67\times {10}^{-15}\mathrm{m} \end{gathered}$$

Its is not a visible wavelength.

Part (e) Step 1: Given information

We have to find the mass of the particle.

Simplify

Since,

$$\begin{gathered} {\mathrm{E}}_4=32.9\mathrm{MeV} \\ {\mathrm{E}}_4=\frac{16\times {(6.62\times {10}^{-34}\mathrm{J}.\mathrm{s})}^2}{2\mathrm{m}\times {10}^{-30}\mathrm{m}} \\ \mathrm{m}\approx {10}^{-27}\mathrm{kg} \end{gathered}$$

This particle can be proton.