Question

An electron is confined in a one-dimensional box of length \(L\). Another electron is confined in a box of length 2 \(L\). Both are in the ground state. What is the ratio of their energies $E_{2 l} / E_{L} ?$

Short answer

Answer: The ratio of the energies of the electrons is \(E_{2L}/E_{L} = \frac{1}{4}\).

Step-by-step solution

Calculate the energy of the electron in a box of length L

Using the formula for the energy of a particle in a one-dimensional box, we can calculate the ground-state energy (\(E_1\)) of an electron in a box of length \(L\): $$ E_{L} = \frac{1^2 \hbar^2 \pi^2}{2mL^2} $$

Calculate the energy of the electron in a box of length 2L

Similarly, we can calculate the ground-state energy (\(E'_1\)) of an electron in a box of length \(2L\): $$ E_{2L} = \frac{1^2 \hbar^2 \pi^2}{2m(2L)^2} $$

Find the ratio of their energies \(E_{2L} / E_{L}\)

Dividing the energy of the electron in the box of length \(2L\) by the energy of the electron in the box of length \(L\), we get: $$ \frac{E_{2L}}{E_{L}} = \frac{\frac{1^2 \hbar^2 \pi^2}{2m(2L)^2}}{\frac{1^2 \hbar^2 \pi^2}{2mL^2}} $$

Simplify the expression

Simplifying the expression, we get: $$ \frac{E_{2L}}{E_{L}} = \frac{\frac{1^2 \hbar^2 \pi^2}{2m(2L)^2}}{\frac{1^2 \hbar^2 \pi^2}{2mL^2}} = \frac{1^2 \hbar^2 \pi^2 \cdot 2mL^2}{2m(2L)^2\cdot 1^2 \hbar^2 \pi^2} = \frac{1}{2^2} = \frac{1}{4} $$ Therefore, the ratio of the energies of the electrons is \(E_{2L}/E_{L} = \frac{1}{4}\).