Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the energy levels for a particle in a rigid box are given by
$$
E_n = \frac{n^2 h^2}{8mL^2}
$$
So, the four lowest energy levels of the electron are
$$E_1 = \frac{(1)^2 (6.63\times 10^{-34})^2}{8(9.11\times 10^{-31})(0.1\times 10^{-9})^2} $$
$$ E_1 =\bf 6.02 \times 10^{-18} \; \rm {J}=\color{red}{ \bf 37.6} \; \rm {eV} $$
$$E_2 = \frac{(2)^2 (6.63\times 10^{-34})^2}{8(9.11\times 10^{-31})(0.1\times 10^{-9})^2} $$
$$ E_2 = \bf2.41 \times 10^{-17} \; \rm {J}=\color{red}{ \bf 150} \; \rm {eV} $$
$$E_3 = \frac{(3)^2 (6.63\times 10^{-34})^2}{8(9.11\times 10^{-31})(0.1\times 10^{-9})^2} $$
$$ E_3 = \bf5.42 \times 10^{-17} \; \rm {J} = \color{red}{ \bf338} \; \rm {eV} $$
$$E_4 = \frac{(4)^2 (6.63\times 10^{-34})^2}{8(9.11\times 10^{-31})(0.1\times 10^{-9})^2} $$
$$ E_4 =\bf 9.64 \times 10^{-17} \; \rm {J} =\color{red}{ \bf 602 }\; \rm {eV}$$
$$\color{blue}{\bf [b]}$$
To find the wavelengths corresponding to the transitions between energy levels, we can use the formula of
$$
\Delta E= E_n-E_m= \frac{hc}{\lambda_{n\rightarrow m}}
$$
So,
$$
\lambda_{n\rightarrow m}= \frac{hc}{E_n-E_m}
$$
Now we need to find the wavelengths for the transitions $ \lambda_{2 \rightarrow 1} $, $ \lambda_{3 \rightarrow 1} $, $ \lambda_{3 \rightarrow 2} $, $ \lambda_{4 \rightarrow 1} $, $ \lambda_{4 \rightarrow 2} $, and $ \lambda_{4 \rightarrow 3} $.
$$
\lambda_{2\rightarrow 1}= \frac{hc}{E_2-E_1}=\frac{(6.63\times 10^{-34})(3\times 10^8)}{(2.41 \times 10^{-17})-(6.02\times 10^{-18})}
$$
$$ \lambda_{2 \rightarrow 1} =\color{red}{ \bf 11.0} \; \rm {nm} $$
By the same approach,
$$ \lambda_{3 \rightarrow 1} =\color{red}{ \bf 4.12} \; \rm {nm} $$
$$ \lambda_{3 \rightarrow 2} =\color{red}{ \bf 6.60} \; \rm {nm} $$
$$ \lambda_{4 \rightarrow 1} =\color{red}{ \bf 2.20 }\; \rm {nm} $$
$$ \lambda_{4 \rightarrow 2} =\color{red}{ \bf 2.75 }\; \rm {nm} $$
$$ \lambda_{4 \rightarrow 3} = \color{red}{ \bf 4.71} \; \rm {nm} $$
$$\color{blue}{\bf [c]}$$
We know that:
$\bullet$ Infrared: $ \lambda > 700 \; \text{nm} $
$\bullet$ Visible light: $ 400 \; \text{nm} \leq \lambda \leq 700 \; \text{nm} $
$\bullet$ Ultraviolet: $ \lambda < 400 \; \text{nm} $
Since all of these wavelengths are in the range of a few nanometers, they fall in the $\bf ultraviolet \;(UV)$ portion of the electromagnetic spectrum.
$$\color{blue}{\bf [d]}$$
In the Bohr hydrogen atom model, energy levels are negative, which indicates that the electron is bound to the nucleus. The energy is negative because the electron is in a bound state, and it would require energy to free the electron. In this rigid box model, the energies are positive, which means the particle is not considered bound in the same way as in the Bohr atom.
This difference is not physically significant; it's just a result of how the models represent the energy levels.
$$\color{blue}{\bf [e]}$$
$\Rightarrow$ Similarity: Both the Bohr model and the rigid box model have quantized energy levels, meaning the electron can only occupy specific, discrete energies.
$\Rightarrow$ Differences: In the Bohr model, the electron undergoes orbital motion around the nucleus while the electron in a rigid box undergoes translational.
In the Bohr model, the energy levels are negative. In the rigid box model, the particle is confined to a box and has positive energy levels. Additionally, the energy levels in the Bohr model get closer together as $n$ increases, while the energy levels get farther apart as $n$ increases in the rigid box model,
