An electron is trapped in a one-dimensional infinite potential well that is 100 pm wide; the electron is in its ground state. What is the probability that you can detect the electron in an interval of width centered at x = (a) 25 pm, (b) 50 pm, and (c) 90 pm? (Hint: The interval x is so narrow that you can take the probability density to be constant within it.)

An electron is a negatively charged subatomic particle. It can be either free (not attached to any atom), or bound to the nucleus of an atom. Electrons in atoms exist in spherical shells of various radii, representing energy levels. The larger the spherical shell, the higher the energy contained in the electron.

Which o the infinite potential well,

L = 100 pm

Width of the interval, $∆E=5.0\mathrm{pm}$

Probability of finding electron in any interval, role="math" localid="1661767181193" $p=\int {{\left|\psi \right|}^2}_{}∆x$

For a small interval,$p={{\left|\psi \right|}^2}_{}∆x$

By substituting the value of

role="math" localid="1661767285650" $$\begin{gathered} {{\left|\psi \right|}^2}_{}=\frac{2}{L}{\sin }^2\left(\frac{\mathrm{\pi x}}{L}\right) \\ Weget∆xp=\left(\frac{2∆x}{L}\right){\sin }^2\left(\frac{\mathrm{\pi x}}{L}\right) \end{gathered}$$

(a)

Here, x = 25 pm

$$\begin{gathered} p=\left(\frac{2∆\mathrm{x}}{\mathrm{L}}\right){\sin }^2\left(\frac{\mathrm{\pi x}}{\mathrm{L}}\right) \\ =\left(\frac{2(5.0\mathrm{pm})}{100\mathrm{pm}}\right){\sin }^2\left(\frac{\mathrm{\pi }(5.0\mathrm{pm})}{100\mathrm{pm}}\right) \\ =0.050 \end{gathered}$$

Hence, the value is

0.050 x .

(b)

Here, x = 50 pm

$p=\left(\frac{2∆x}{L}\right){\sin }^2\left(\frac{\mathrm{\pi x}}{L}\right)$

Hence,

$$\begin{gathered} p=\left(\frac{2∆x}{L}\right){\sin }^2\left(\frac{\mathrm{\pi x}}{L}\right) \\ =\left(\frac{2\left(5.0pm\right)}{100pm}\right){\sin }^2\left(\frac{\mathrm{\pi }\left(5.0pm\right)}{100pm}\right) \\ =0.10 \end{gathered}$$

Hence, the value is 0.10pm .

(c)

Here, x = 90 pm

$p=\left(\frac{2∆x}{L}\right){\sin }^2\left(\frac{\mathrm{\pi x}}{L}\right)$

Hence,

$$\begin{gathered} p=\left(\frac{2∆x}{L}\right){\sin }^2\left(\frac{\mathrm{\pi x}}{L}\right) \\ =\left(\frac{2\left(5.0\mathrm{pm}\right)}{100\mathrm{pm}}\right){\sin }^2\left(\frac{\mathrm{\pi }(90\mathrm{pm})}{100\mathrm{pm}}\right) \\ =0.0095 \end{gathered}$$

Hence, the value is

0.0095 pm