Answer
$L = 8.5\times 10^{-10}~m$
Work Step by Step
We can express the energy in units of joules:
$E = (4.7~eV)(\frac{1.6\times 10^{-19}~J}{1~eV}) = 7.52\times 10^{-19}~J$
We can find the width of the potential well:
$E_n = (\frac{h^2}{8m~L^2})~n^2$
$E_3 = (\frac{h^2}{8m~L^2})~(3)^2$
$E_3 = \frac{9~h^2}{8m~L^2}$
$L^2 = \frac{9~h^2}{8m~E_3}$
$L = \frac{3~h}{\sqrt{8m~E_3}}$
$L = \frac{(3)~(6.626\times 10^{-34}~J~s)}{\sqrt{(8)(9.109\times 10^{-31}~kg)~(7.52\times 10^{-19}~J)}}$
$L = 8.5\times 10^{-10}~m$
