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, E_{0}=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 E_{1}=(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, E_{1}=5/2h(bar)=3.3eV=3.3x1.6x10^{-19}J

from this we find =2.011x10^{15}rad/s

for 2nd excited state, n_{x}=n_{y}=1,
n_{z}=0

therefore, E_{2}=(1+1+3/2)h(bar)=7/2h(bar)=7/2x1.05x10^{-34}x2.011x10^{15}=7.39x10^{-19}J=4.61eV

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.

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

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