Question

It takes 4.0 eV to excite an electron in a box from the ground level to the first excited level. What is the width L of the box?

Answer 1

To solve this problem we will use the concept related to electrons in a box which determines the energy of an electron in that state.

Mathematically this expression is given by,

$E_n= \frac{n^2h^2}{8mL^2}$

Where,

m = mass of an electron

h = Planck's constant

n = is the integer number of the eigenstate

L = Quantum well width

The change in energy must be given in state 1 and 2, therefore

$\Delta E = E_2 - E_1$

$\Delta E = \frac{2^2h^2}{8mL^2}-\frac{1^2h^2}{8mL^2}$

$\Delta E = \frac{3h^2}{8mL^2}$

Replacing we have:

$(4*1.6*10^{-19}) = \frac{3(6.626*10^{34})}{8*(9.11*10^{-31})*L^2}$

$L = 0.53nm$

Therefore the correct answer is C.