# Answer on Quantum Mechanics Question for JOEL LOBO

Question #2149

A quantum particle confined to one dimensional box of width ‘a’ is in its first

exited state. What is the probability of finding the particle over an interval of

(a/2) marked symmetrically at the centre of the box.

exited state. What is the probability of finding the particle over an interval of

(a/2) marked symmetrically at the centre of the box.

Expert's answer

The probability density function for the particle in one dimensional box in its first exited state is

<img src="/cgi-bin/mimetex.cgi?%5Cpsi%20%5Ccdot%20%5Cpsi*%20=%20%5Cfrac%7B2%7D%7Ba%7D%20%5Csin%5E2%7B%5Cfrac%7B2%20%5Cpi%20x%7D%7Ba%7D%7D" title="\psi \cdot \psi* = \frac{2}{a} \sin^2{\frac{2 \pi x}{a}}">

It becomes equal to zero at x = a/2.

<img src="/cgi-bin/mimetex.cgi?%5Cpsi%20%5Ccdot%20%5Cpsi*%20=%20%5Cfrac%7B2%7D%7Ba%7D%20%5Csin%5E2%7B%5Cfrac%7B2%20%5Cpi%20x%7D%7Ba%7D%7D" title="\psi \cdot \psi* = \frac{2}{a} \sin^2{\frac{2 \pi x}{a}}">

It becomes equal to zero at x = a/2.

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