Question #188432

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


1
Expert's answer
2021-05-06T07:40:21-0400

Answer

Energy for ground state

E=h28mL2E=(6.63×1034)28(9.1×1031)(1010)2=37.737eVE=\frac{h^2}{8mL^2}\\E=\frac{(6.63\times10^{-34})^2}{8(9.1\times10^{-31})(10^{-10})^2}\\ =37.737eV



E2=4E=150.95eVE_2=4E=150.95eV


E3=9E=339.639eVE_3=9E=339.639eV


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