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
Energy for ground state
E=h28mL2E=(6.63×10−34)28(9.1×10−31)(10−10)2=37.737eVE=\frac{h^2}{8mL^2}\\E=\frac{(6.63\times10^{-34})^2}{8(9.1\times10^{-31})(10^{-10})^2}\\ =37.737eVE=8mL2h2E=8(9.1×10−31)(10−10)2(6.63×10−34)2=37.737eV
E2=4E=150.95eVE_2=4E=150.95eVE2=4E=150.95eV
E3=9E=339.639eVE_3=9E=339.639eVE3=9E=339.639eV
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