Question

Consider the electron states in an infinite square well potential.
a) If the difference in energy between the n=2 and the n=3 states is 2 eV, calculate the width
of this square well.
b) If energy making a transition from the n=3 state to the n=2 state gives up the energy
difference as an emitted photon, what is the wavelength of the photon?

Answer 1

Energy levels of infinite square well potential is $E=\frac{n^2h^2}{8ma^2}$

a) For $n=2$ state, $E_2=\frac{4h^2}{8ma^2}$

For $n=3$ state, $E_3=\frac{9h^2}{8ma^2}$

Difference in energy levels $E_3-E_2=2\,eV=2*1.6*10^{-19}\,J$

$E_3-E_2=\frac{9h^2}{8ma^2}-\frac{4h^2}{8ma^2}=\frac{5h^2}{8ma^2}$

$a=\sqrt{\frac{5h^2}{8m(E_3-E_2)}}$

$a=\sqrt{\frac{5(6.626*10^{-34})^2}{8(9.11*10^{-31})(2*1.6*10^{-19})}}$

$a=0.97*10^{-9}\,m=0.97\,nm$

Width of square well is $a=0.97*10^{-9}\,m=0.97\,nm$

b)

$E_3-E_2=\frac{hc}{\lambda}$

$\lambda=\frac{hc}{E_3-E_2}$

$\lambda=\frac{6.626*10^{-34}*3*10^8}{2*1.6*10^{-19}}=621.2*10^{-9}\,m=621.2\,nm$

Wavelength of photon is $\lambda=621.2\,nm$