Question

It takes 3.0 eV of energy to excite an electron in a 1-dimensional infinite well from the
ground state to the first excited state. What is the width, L, of the box?

Answer 1

For an infinite potential well, the energy of a particular energy level n, where n = 1, 2, 3... is given by \[E_{n}=\frac{\hbar^{2}\pi^{2}n^{2}}{2mL^{2}}\]where L is the width of the well.
The ground state corresponds to n = 1 and the 1st excited state corresponds to n = 2. therefore teh difference in energy between teh two levels is
\[E_{2}-E_{1}=\frac{\hbar^{2}\pi^{2}2^{2}}{2mL^{2}}-\frac{\hbar^{2}\pi^{2}1^{2}}{2mL^{2}}\] and hence
\[E_{2}-E_{1}=\frac{4\hbar^{2}\pi^{2}}{2mL^{2}}-\frac{\hbar^{2}\pi^{2}}{2mL^{2}}\] meaning that
\[E_{2}-E_{1}=\frac{3\hbar^{2}\pi^{2}}{2mL^{2}}\]
we know that the energy of transition is 3eV = (E_{2} -E_{1}), which we must convert to Joules (by multiplying the value by the elementary electronic charge). Also rearranging we get \[L^{2}=\frac{3\hbar^{2}\pi^{2}}{2m(E_{2}-E_{1})}\] and hence \[L=\sqrt{\frac{3\hbar^{2}\pi^{2}}{2m(E_{2}-E_{1})}}\]
Plugging in the numbers we get
\[L=\sqrt{\frac{3\hbar^{2}\pi^{2}}{2m(3\times1.602\times10^{-19})}}\](note that's a -19 in the bottom line, as it doesn't come out too clear)
\[L=6.132\times10^{-10}\textrm{ metres}\]or L is equal to 0.6132 nanometres or 6.132 Angstroms (which ever sets of units you wish to use).