Question

A quantum mechanical simple harmonic oscillator has a 1st excited state with energy 3.3 eV and there are eight spin-1/2 particles in the oscillator. How much energy is needed to add a ninth electron. Explain and show your work

Answer 1

for Nth excited state energy of a 3-D simple harmonic oscillator,E_n= (n_x+n_y+n_z+3/2)h(bar)

for ground state we consider n_x=n_y=n_z=0

hence, E₀=3/2h(bar)=3x1/2h(bar), which means it can accomodate 3 spin half electrons

next for 1st excited state, n_x=1, n_y=n_z=0

hence E₁=(1+3/2)h(bar)=5/2h(bar)=5x1/2h(bar), which means it can accomodate 5 spin half electrons

total=3+5=8 electrons.

so the ninth electron must go to the third excited state

now, E₁=5/2h(bar)=3.3eV=3.3x1.6x10^-19J

from this we find =2.011x10¹⁵rad/s

for 2nd excited state, n_x=n_y=1, n_z=0

therefore, E₂=(1+1+3/2)h(bar)=7/2h(bar)=7/2x1.05x10^-34x2.011x10¹⁵=7.39x10^-19J=4.61eV