Question

An electron is in the 3d state of a hydrogen atom. The most probable distance of the electron from the proton is$\mathbf{9}{\mathit{a}}_{\mathbf{o}}$. What is the probability that the electron would be found between$\mathbf{8}{\mathit{a}}_{\mathbf{o}}\mathit{a}\mathit{n}\mathit{d}\mathbf{10}{\mathit{a}}_{\mathbf{o}}$?

Short answer

The probability of finding the electron between$8a_{o}and10a_{o}is0.212.$

Step-by-step solution

 Given data

The state of an electron – 3d.

 Concept

The space around the proton in a hydrogen atom, in which the probability of finding the electron is at least $\mathbf{90}\mathbf{\%}$, is known as the orbital.

 Solution

The radial function is given as-

$R_{3,2}\left(r\right)=\frac{1}{{\left(3a_o\right)}^{3/2}}\frac{\left(2\sqrt{2r^2}\right)}{\left(27\sqrt{5a_o^2}\right)}{e}^{-r/\left(3a_o\right)}$

The probability to find the electron between and

$$\begin{gathered} P={\int }_{8a_o}^{10a_o}{R_{3,2}}^2\left(r\right)r^{2}dr \\ ={\int }_{8a_o}^{10a_o}{\left(\frac{1}{{\left(3a_{°}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\frac{2\sqrt{2}{r}^2}{27{\sqrt{5}}_0^2}{e}^{-r/3a_{°}}\right)}^2{r}^{2}dr \\ =\frac{1}{{\left(3a_{°}\right)}^3}\frac{8}{{27}^2\times 5a_0^4}{\int }_{8a_o}^{10a_o}{r}^6{e}^{-r/3a_{°}}dr \end{gathered}$$

Let $\frac{r}{a_o}=x$ , then the equation becomes-

$$\begin{gathered} P=\frac{8}{{27}^3\times 5}{\int }_{8a_o}^{10a_o}{X}^6{e}^{2x/3}dr \\ =0.212 \end{gathered}$$

This integral is found by the calculator.

Therefore, the probability of finding the electron between$8a_{o}and10a_{o}is0.212.$