91Ó°ÊÓ

The given data is listed below.

The radius of the electron is greater than the Bohr radius.

The formula for finding the probability of electron in the ground state of hydrogen atom inside a sphere of radius r is given by,

$\mathbf{p}\left(r\right)\mathbf{=}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{a}}\left(1+2x+2x^2\right)$

Here, x = 1 and r = a .

Here, a is the Bohr radius.

The probability of finding the electron in the ground state of a hydrogen atom found inside a sphere of radius r is given by-

$P\left(r\right)=\left[1-e^{-2x}\left(1+2x+2x^2\right)\right]$

Here, x = na and a is the Bohr radius.

For, r = a and x = 1 .

$$\begin{gathered} P\left(a\right)=\left[1-e^{-2}\left(1+2+2\right)\right] \\ =1-5e^{-2} \\ =1-\left(5×0.135\right) \\ =0.323 \end{gathered}$$

Now, the probability that the electron can be found outside this sphere is:

$$\begin{gathered} P=1-0.322 \\ =0.677 \end{gathered}$$

$$\begin{gathered} P\%=\left(0.677×100\right)\% \\ =68\% \end{gathered}$$

Thus, the probability the electron will be found at a radius greater than the Bohr radius is 68% .