# Question f02d4

Mar 13, 2017

First part as below.
Bullet part can not be calculated as explained below.

#### Explanation:

It is to be noted that the function ${\psi}_{n} \left(x\right)$ is normalized wave function. Also $n$ can take values as $1 , 2 , 3. \ldots .$ for ground and higher excited states.

Pictorial representation is as below (read $l$ for L in the figure). For first excited state $n = 2$
Hence, Probability of finding the electron at a point $x$ within the width $l$ of infinite potential well for $n = 2$ is
$\psi {\text{*}}_{2} \left(x\right) {\psi}_{2} \left(x\right) = \frac{2}{l} {\sin}^{2} \left(\frac{2 \pi}{l} x\right)$ ......(1)

Probability of finding electron in the first excited state between the values $x = 0 \mathmr{and} x = 0.20 \times {10}^{-} 10 m$ is definite integral of above equation with respect to variable $x$ between the given limits.

$P = {\int}_{0}^{0.2 \times {10}^{-} 10} \psi {\text{*}}_{2} \left(x\right) {\psi}_{2} \left(x\right) \mathrm{dx}$
$\implies P = {\int}_{0}^{0.2 \times {10}^{-} 10} \frac{2}{l} {\sin}^{2} \left(\frac{2 \pi}{l} x\right) \mathrm{dx}$

Using the half angle formula,
${\sin}^{2} \left(x\right) = \frac{1}{2} \cdot \left(1 - \cos \left(2 x\right)\right)$ and substituting into the indefinite integral we have
$P = \frac{2}{l} \int \left[\frac{1}{2} \left(1 - \cos \left(2 \frac{2 \pi}{l} x\right)\right)\right] \mathrm{dx}$
=>P=1/lint(1-cos((4pi)/l x)dx#
$\implies P = \frac{1}{l} \left[\int \mathrm{dx} - \int \cos \left(\frac{4 \pi}{l} x\right) \mathrm{dx}\right]$ ......(2)
Now integrating both terms and applying the limits we get

$P = \frac{1}{l} {\left[x - \frac{\sin \left(\frac{4 \pi}{l} x\right)}{\frac{4 \pi}{l}}\right]}_{0}^{0.2 \times {10}^{-} 10}$
$\implies P = \frac{0.2 \times {10}^{-} 10}{l} - \frac{1}{4 \pi} \sin \left(\frac{0.8 \times {10}^{-} 10 \pi}{l}\right)$
is the required expression in terms of width $l$ of the infinite potential well.

As value of width $l$ has not been given, the numerical value of Probability of finding electron in the first excited state between the values $x = 0.20 \times {10}^{-} 10 m \mathmr{and} x = 1.00 \times {10}^{-} 10 m$ can not be calculated.