91Ó°ÊÓ

A particle is in the first excited state, $n=2$.

The wavefunction for an infinite well is given by

localid="1657615959392" $\mathrm{ψ}n\left(x\right)=2L\sin \left(n\mathrm{π}xL\right)  \left(1\right)$

Where $\mathit{n}$is the number of energy states and localid="1657616755075" $\mathit{L}$is the width of an infinite well

The expression for probability is given by

$Pn\left(\mathrm{x}\right)=\left|\mathrm{ψ}\mathrm{n}\left(\mathrm{x}{)}^{*}\mathrm{ψ}\mathrm{n}\right(\mathrm{x})\right|\left(2\right)$

for the first excited state, $n=2$.

Substitute the value of n in equation (1)

localid="1659187597446" $\mathrm{ψ}n\left(x\right)=2L\sin \left(2\mathrm{π}xL\right)$

Hence, the probability is

$\mathrm{P}2\left(x\right)=\left|\mathrm{Ψ}2\left(x{)}^{*}\mathrm{Ψ}2\right(x)\right|=2L\sin 2\left(2\mathrm{π}xL\right)$

To find the most probable position of a particle in the energy state, maximize the probability $P2\left(x\right)$That is,

$$\begin{gathered} \mathrm{dP}2\left(x\right)dx=2(2\mathrm{L})\sin (2\mathrm{Ï€}x\mathrm{L})\cos (2\mathrm{Ï€}x\mathrm{L})\left(2\mathrm{Ï€}\mathrm{L}\right) \\ =2\mathrm{Ï€}\left(4L2\right)\sin \left(2\mathrm{Ï€}xL\right)\cos \left(2\mathrm{Ï€}xL\right) \end{gathered}$$

The quantity$dP2\left(x\right)dx=0$

when$x=0,L4,L2,3L4,L$.

But when$x=0,L2,L$

the wavefunction$\mathrm{ψ}2\left(x\right)=0$.

Thus, the probability of finding the particle is zero.

$x=0,L2,L$