Answer
$\frac{L'}{L} = 1.41$
Work Step by Step
In the ground state, $n=1$
We can write an expression for the ground-state energy of a particle in a potential well:
$E_1 = (\frac{h^2}{8m~L^2})~n^2 = \frac{h^2}{8m~L^2}$
We can find $\frac{L'}{L}$:
$E_1' = (\frac{h^2}{8m~L'^2})~n^2$
$E_1' = \frac{h^2}{8m~L'^2} = 0.500~\frac{h^2}{8m~L^2}$
$\frac{1}{L'^2} = \frac{0.500}{L^2}$
$\frac{L'^2}{L^2} = \frac{1}{0.500}$
$\frac{L'^2}{L^2} = 2.00$
$\frac{L'}{L} = \sqrt{2.00}$
$\frac{L'}{L} = 1.41$
