91Ó°ÊÓ

The width of the one-dimensional infinite potential well is $L$.

The expression to calculate the wave function for the electron is given as follows.

${\mathit{Ψ}}_{\mathbf{n}}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{L}}}\mathit{s}\mathit{i}\mathit{n}\left(\frac{n\mathrm{Ï€}}{L}x\right)$ …(¾±)

The probability of detecting the electron is given by,

$P_n=|{\mathrm{Ψ}}_n{|}^2$

Substitute $\sqrt{\frac{2}{L}}\sin \left(\frac{n\mathrm{π}}{L}x\right)$for ${\mathrm{Ψ}}_n$into above equation.

$P_n=\frac{2}{L}{sin}^2\left(\frac{n\mathrm{Ï€}}{L}x\right)$ …(¾±¾±)

Initially electron is in the ground state, therefore$n=1$

Substitute$1$for $n$into equation (ii).

$P_1=\frac{2}{L}{sin}^2\left(\frac{\mathrm{Ï€}}{L}X\right)$

When electron is excited from ground state to first state excited, then $n=2$.

Substitute2 for $n$into equation (ii).

$P_2=\frac{2}{L}{sin}^2\left(\frac{2\mathrm{Ï€}}{L}X\right)$

To check the probability, divide the probability $P_1$by $P_2$.

$$\begin{gathered} \frac{P_2}{P_1}=\frac{{\displaystyle \frac{2}{L}}{\sin }^2\left({\displaystyle \frac{2\mathrm{Ï€}}{L}}x\right)}{\frac{2}{L}{\sin }^2\left(\frac{2\mathrm{Ï€}}{L}x\right)} \\ \frac{P_2}{P_1}=\frac{{\sin }^2\left(\frac{2\mathrm{Ï€}}{L}x\right)}{{\sin }^2\left(\frac{2\mathrm{Ï€}}{L}x\right)} \end{gathered}$$ …(¾±¾±¾±)

Let assume point near the centre,$x=0.51L$ .

$$\begin{gathered} \frac{P_2}{P_1}=\frac{{\sin }^2\left(\frac{2\mathrm{π}}{L}×0.51L\right)}{{\sin }^2\left(\frac{2\mathrm{π}}{L}×0.51L\right)} \\ \frac{P_2}{P_1}=\frac{{\sin }^2\left(1.01\mathrm{π}\right)}{{\sin }^2\left(1.51\mathrm{π}\right)} \\ \frac{P_2}{P_1}<1 \end{gathered}$$

From the above discussion, the probability is decreases as the electron go from the ground state to first excited state at the centre.

Let assume the point near the well walls,$x=0.99L$

Substitute$0.99L$ for $x$into equation (iii).

$$\begin{gathered} \frac{P_2}{P_1}=\frac{{\sin }^2\left(\frac{2\mathrm{π}}{L}×0.99L\right)}{{\sin }^2\left(\frac{2\mathrm{π}}{L}×0.99L\right)} \\ \frac{P_2}{P_1}=\frac{{\sin }^2\left(1.98\mathrm{π}\right)}{{\sin }^2\left(0.99\mathrm{π}\right)} \\ \frac{P_2}{P_1}>1 \end{gathered}$$

From the above discussion, the probability is increases as the electron go from the ground state to first excited near the one of the well walls.