Answer to Question #188432 in Quantum Mechanics for Haa

Question

An electron is confined in a cubical box of each side 1 Å. Calculate the energies of the electron in

the ground state and the first excited state

Answer 1

Answer

Energy for ground state

$E=\frac{h^2}{8mL^2}\\E=\frac{(6.63\times10^{-34})^2}{8(9.1\times10^{-31})(10^{-10})^2}\\ =37.737eV$

$E_2=4E=150.95eV$

$E_3=9E=339.639eV$