Answer to Question #218093 in Quantum Mechanics for Mohammed Ariful

Question

** Estimate the energy splitting between the lowest two levels for an electron in a three-dimensional cube-shaped potential box with the side length 10 nm.

1◦ 1 meV

2◦ 10 meV

3◦ 100 meV

4◦ 0.11 eV

5◦ 0.22 eV

6◦ 1.12 eV

Answer 1

Energy in ground state, $E_1=\dfrac{(1)^2\pi^2\hbar^2}{2m_eL^2}$

$E_1=\dfrac{\pi^2(1.054\times10^{-34})^2}{2(9.1\times10^{-31})(10\times10^{-9})^2}$

$E_1=6.022\times10^{-22}\space J$

Energy in first excited state, $E_2=\dfrac{(2)^2\pi^2\hbar^2}{2m_eL^2}=4\times6.022\times10^{-22}\space J$

Energy difference, $\Delta E=E_2-E_1=3\times6.022\times10^{-22}\space J$

$\Delta E=112.82\times10^{-3}\space eV=0.11\space eV$