Answer
$0.02$
Work Step by Step
The normalized wavefunction $\psi_n(x)$ for an electron in an infinite, one-dimensional potential well with length $L$ along an $x$ axis are given by
$\psi_n(x)=\sqrt {\frac{2}{L}}\sin\Big(\frac{n\pi}{L}x\Big),\;\;\;\text{for}\;n=1,2,3....,$
where $n$ is the quantum number.
Now, the probability that the electron will be detected in the interval between $x$ and $x+dx$ inside of this well is given by
$p(x)=\psi^2_n(x)dx$
or, $p(x)={\frac{2}{L}}\sin^2\Big(\frac{n\pi}{L}x\Big)dx$
Here, the electron in the potential well in its third excited state $(n=4)$. The electron detector has been centered on a point of maximum probability density.
If we plot the probability density for the third excited state with respect to $x$, we will see that the maximum probability occurs at $x=\frac{L}{8},\frac{3L}{8},\frac{5L}{8},\frac{7L}{8} $. This implies the maximum probability occurs at $x=\frac{(2n^\prime+1)L}{8}$, where $n^\prime=0,1,2,3$
The given infinite well has a length $(L)$ of $200\;pm$ and the detector has a probe of width $2.00\;pm$ $(\text{i.e,}\;dx=2.00\;pm)$
Thus for third excited state, the maximum probabily becomes
$p\Big(\frac{(2n^\prime+1)L}{8}\Big)={\frac{2}{200\;pm}}\sin^2\Big(\frac{4\pi(2n^\prime+1)}{8}\Big)(2\;pm)$
or, $p\Big(\frac{(2n^\prime+1)L}{8}\Big)={\frac{4}{200}}\sin^2\Big(\frac{\pi(2n^\prime+1)}{2}\Big)$
For, $n^\prime=0,1,2,3$; $\sin^2\Big(\frac{\pi(2n^\prime+1)}{2}\Big)=1$
Therefore,
$p\Big(\frac{(2n^\prime+1)L}{8}\Big)={\frac{4}{200}}=0.02$
$\therefore$ The probability of detection by the probe is $0.02$
