For an infinite square well of length L,what is the ratio of energy difference between the ground and first excited state for a square well of length L to that of 2 L?
A)0.5
B)1
C)2
D)4
E)5

Question

Quizplus · Accepted Answer

The energy levels for an infinite square well are given by $E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$. The energy difference between the ground state (n=1) and the first excited state (n=2) for a well of length L is $\Delta E_L = E_2 - E_1 = \frac{3 \pi^2 \hbar^2}{2mL^2}$. For a well of length 2L, the energy difference is $\Delta E_{2L} = \frac{3 \pi^2 \hbar^2}{8mL^2}$. The ratio is $\frac{\Delta E_L}{\Delta E_{2L}} = 4$.

For an Infinite Square Well of Length L,what Is the Ratio