Question

# A quantum mechanical simple harmonic oscillator has a 1st excited state with energy 3.3 eV and...

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

for Nth excited state energy of a 3-D simple harmonic oscillator,En= (nx+ny+nz+3/2)h(bar)

for ground state we consider nx=ny=nz=0

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

next for 1st excited state, nx=1, ny=nz=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

for 2nd excited state, nx=ny=1, nz=0

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

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