Where would a particle in the first excited state (first above ground) of an infinite well most likely be found?

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

The wavefunction for an infinite well is given by

localid="1657615959392" $\psi n\left(x\right)=2L\sin \left(n\pi xL\right)\hspace{1em}\hspace{1em}\left(1\right)$

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

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

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

Substitute the value of n in equation (1)

localid="1659187597446" $\psi n\left(x\right)=2L\sin \left(2\pi xL\right)$

Hence, the probability is

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

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

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

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

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

But when$x=0,L2,L$

the wavefunction$\psi 2\left(x\right)=0$.

Thus, the probability of finding the particle is zero.

$x=0,L2,L$