Question #177723

Consider an infinitely deep potential well of width 2a centered at x = 0.

The probability of finding a particle of mass m in the ground, first and

second excited state is 60%, 30% and 10% respectively. What is the energy

expectation value of the particle ?


1
Expert's answer
2021-04-05T11:11:30-0400

Permitted energy level for nth state, E=n2h28ml2E=\dfrac{n^{2}h^{2}}{8ml^2}

For Ground state, n=1

E1=12×(4.135×1015)28×9.1×1031×4a2E1=0.587a2eVE_1=\dfrac{1^2\times(4.135\times10^{-15})^2}{8\times9.1\times10^{-31}\times4a^2}\\ E_1=\dfrac{0.587}{a^2}eV


For First Excited state, n=2

E2=(2)2×0.587a2=2.34a2eVE_2=\dfrac{(2)^2\times0.587}{a^2}=\dfrac{2.34}{a^2}eV


For Second Excited state, n=3

E3=9×0.587a2=5.283a2eVE_3=\dfrac{9\times0.587}{a^2}=\dfrac{5.283}{a^2}eV


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