Answer to Question #323004 in Quantum Mechanics for Bhargavi

Question

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

Answer 1

Answer

Width l=10$A°$

Interval

$x=5A°$

Since the electron is in its least energy state,

n=1

So probabilty is

$\psi_n=\sqrt{\frac{2}{l}}sin(\frac{n\pi x}{l})$

Then probabilty

$\psi_n=\sqrt{\frac{2}{10}}sin(\frac{\pi 5}{10}) \\=0.45$