Question #323004

A particle is moving in a one dimensional width 25 Armstrong assuming the particle is in its least energy state calculate the probability of finding particle in an interval of 5 Armstrong at the distances of A/2,A/3 and at A where a is the width of the potential well


1
Expert's answer
2022-04-04T09:04:00-0400

Answer

Width l=10A°

Interval

x=5A°x=5A°

Since the electron is in its least energy state,

n=1

So probabilty is

ψn=2lsin(nπxl)\psi_n=\sqrt{\frac{2}{l}}sin(\frac{n\pi x}{l})

Then probabilty

ψn=210sin(π510)=0.45\psi_n=\sqrt{\frac{2}{10}}sin(\frac{\pi 5}{10}) \\=0.45





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