Answer
The new ground-state energy is $~~0.65~eV$
Work Step by Step
In the ground state, $n=1$
We can write an expression for the ground-state energy of an electron in a potential well:
$E_1 = (\frac{h^2}{8m~L^2})~n^2 = \frac{h^2}{8m~L^2}$
We can write an expression for the ground-state energy of an electron in a potential well if the width $L$ is doubled:
$E_1 = \frac{h^2}{8m~(2L)^2}= \frac{1}{4}~\frac{h^2}{8m~L^2}$
The energy of the ground state of the trapped electron is multiplied by $\frac{1}{4}$
We can find the new ground-state energy:
$E = (\frac{1}{4})(2.6~eV) = 0.65~eV$
The new ground-state energy is $~~0.65~eV$
