A particle of mass $m$ is confined in a one-dimensional box of length $L$. The ratio of the energy difference between the third and second excited states to the energy difference between the second and first excited states is:

Solution

1. The energy levels for a particle in a one-dimensional box are given by $E_n=n^2{E}_1$, where $n=1,2,3,\dots$.
2. The first, second, and third excited states correspond to $n=2$, $n=3$, and $n=4$, respectively.
3. The energy difference between the third and second excited states is $$\begin{array}{rl}{E}_4-E_3& =(4^2-3^2)E_1\\ & =(16-9)E_1\\ & =7E_1\end{array}$$.
4. The energy difference between the second and first excited states is $$\begin{array}{rl}{E}_3-E_2& =(3^2-2^2)E_1\\ & =(9-4)E_1\\ & =5E_1\end{array}$$.
5. The ratio of these energy differences is $\frac{7E_1}{5E_1}=\frac{7}{5}$.
6. Hence the answer is (B).