Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The energy levels for a particle in a one-dimensional box are given by
$$
E_n = \frac{n^2 h^2}{8mL^2}\tag 1
$$
To find $n$ and $n+1$, we will set up two equations for the energy levels using the known values of $ E_n $ and $ E_{n+1} $, then solve for $n$.
We will express the ratio of the two energy levels:
$$
\frac{E_{n+1}}{E_n} = \frac{(n+1)^2}{n^2}
$$
Plug the known;
$$
\frac{51.4}{32.9} = \frac{n^2+2n+1}{n^2}
$$
Thus, $n=\color{red}{\bf 4}$, and hence, $n+1=\color{red}{\bf 5}$
$$\color{blue}{\bf [b]}$$
First, we need to find the energy levels magnitudes from 1 to 5, where we know that $ E_4 = 32.9 \; \rm {MeV} $ and $ E_5 = 51.4 \; \rm {MeV} $
Recalling that
$$E_n=n^2E_1$$
So,
$$E_1=\dfrac{E_n}{n^2}=\dfrac{E_4}{4^2}=\dfrac{32.9 }{16}=\color{red}{\bf 2.06}\;\rm eV$$
Thus,
$$E_2=(2)^2E_1=4(2.06)=\color{red}{\bf 8.23}\;\rm eV$$
$$E_3=(3)^2E_1=9(2.06)=\color{red}{\bf 18.5}\;\rm eV$$
$$E_4= \color{red}{\bf 32.9 }\;\rm eV$$
$$E_5= \color{red}{\bf 51.4 }\;\rm eV$$
$$\color{blue}{\bf [c]}$$
see the graph below.
$$\color{blue}{\bf [d]}$$
The energy difference between the levels $ n + 1 $ and $ n $ is given by
$$
\Delta E =E_{\rm photon}= E_{n+1} - E_n =E_5-E_4$$
where $E_{\rm photon}=\dfrac{hc}{\lambda}$, so
$$\dfrac{hc}{\lambda}=E_5-E_4 $$
Hence, the wavelength of the emitted photon is given by
$$
\lambda = \frac{hc}{E_5-E_4 }
$$
Plug the known;
$$
\lambda = \frac{(6.63\times 10^{-34})(3\times 10^8)}{ (51.4-32.9)(1.6\times 10^{-19}) }=\color{red}{\bf 6.72\times 10^{-8}}\;\rm m
$$
The wavelength of the photon emitted during the $ n + 1 \to n $ transition is approximately $ 6.72 \times 10^{-5} \; \rm {nm} $, which is much shorter than the typical visible light wavelength (which ranges from about 400 nm to 700 nm).
Therefore, this photon is in the gamma-ray region of the electromagnetic spectrum.
$$\color{blue}{\bf [e]}$$
We can determine the mass of the particle using the energy level formula:
$$
E_n = \frac{n^2 h^2}{8 m L^2}
$$
So,
$$
m= \frac{n^2 h^2}{8 L^2E_n }
$$
Solving for $ n =5$,
$$
m= \frac{(5)^2 (6.63\times 10^{-34})^2}{8 (10\times 10^{-15})^2(51.4\times 1.6\times 10^{-19})}=\color{red}{\bf 1.67 \times 10^{-21}} \; \rm {kg}
$$
The particle's mass is not a mass of a proton or even an electron.
