Chapter 1, Problem 1D.7E

(a)

Interpretation Introduction

Interpretation:

The probability of finding an electron anywhere inside a sphere of radius $a_0$ in the ground state of hydrogen atom has to be determined.

Concept Introduction:

A node is an area where the probability of finding the electron is zero.  On the other hand, an atomic orbital refers to an area which is used to describe the wave-like behavior of either one electron or of the electron pair. The square of the wavefunction is used to find the probability of finding the electron around the nucleus in an atom.

Answer to Problem 1D.7E

The probability of finding an electron anywhere inside a sphere of radius $a_0$ in the ground state of hydrogen atom is $0.32.$

Explanation of Solution

The probability density of an electron at a point, when it is in a $1s-$ orbital is given below.

    ${\psi }^2=\frac{1}{\pi a_0^3}{e}^{\frac{-2r}{a_0}}$

Therefore the probability to locate an electron anywhere inside a sphere of radius $a_0$ in the ground state of hydrogen atom can be calculated as follows,

    $\begin{array}{c}\text{Probability}={\displaystyle \underset{0}{\overset{a_0}{\int }}{\psi }^2}dV\\ \\ ={\displaystyle \underset{0}{\overset{a_0}{\int }}\left(\frac{1}{\pi a_0^3}{e}^{\frac{-2r}{a_0}}\right)4\pi r^{2}dr}\\ \\ =\left(\frac{4}{a_0^3}\right){\displaystyle \underset{0}{\overset{a_0}{\int }}{r}^2.e^{\frac{-2r}{a_0}}dr}\\ \\ =\left(\frac{4}{a_0^3}\right){\left[\frac{e^{\frac{-2r}{a_0}}}{\frac{-2}{a_0}}\left(r^2-\frac{2r}{\frac{-2}{a_0}}+\frac{2}{{\left(\frac{-2}{a_0}\right)}^2}\right)\right]}_0^{a_0}\\ \\ =\left(\frac{4}{a_0^3}\right)\left[\frac{e^{\frac{-a_{0}r}{a_0}}}{\frac{-2}{a_0}}\left(a_0^2-\frac{2a_0}{\frac{-2}{a_0}}+\frac{2}{{\left(\frac{-2}{a_0}\right)}^2}\right)-\left(\frac{1}{\frac{-2}{a_0}}\times \frac{2}{{\left(\frac{-2}{a_0}\right)}^2}\right)\right]\\ \\ =\left(\frac{4}{a_0^3}\right)\left[-\frac{a_0^{}{e}^{-2}}{2}\left(a_0^2+a_0^2+\frac{a_0^2}{2}\right)+\frac{a_0^3}{4}\right]\\ \\ =\left(\frac{4}{a_0^3}\right)\left[-a_0^3{e}^{-2}+\frac{a_0^3{e}^{-2}}{4}+\frac{a_0^3}{4}\right]\\ \\ =\left(\frac{4}{a_0^3}\right)\left[\frac{a_0^3-5a_0^3{e}^{-2}}{4}\right]\\ \\ =1-5e^{-2}\\ \\ =0.32.\end{array}$

The probability is $0.32.$

(b)

Interpretation Introduction

Interpretation:

The probability of finding an electron anywhere inside a sphere of radius $2a_0$ in the ground state of hydrogen atom has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 1D.7E

The probability of finding an electron anywhere inside a sphere of radius $a_0$ in the ground state of hydrogen atom is $0.76.$

Explanation of Solution

The probability density of an electron at a point, when it is in a $1s-$ orbital is given below.

    ${\psi }^2=\frac{1}{\pi a_0^3}{e}^{\frac{-2r}{a_0}}$

Therefore the probability to locate an electron anywhere inside a sphere of radius $a_0$ in the ground state of hydrogen atom can be calculated as follows,

    $\begin{array}{c}\text{Probability}={\displaystyle \underset{0}{\overset{2a_0}{\int }}{\psi }^2}dV\\ \\ ={\displaystyle \underset{0}{\overset{2a_0}{\int }}\left(\frac{1}{\pi a_0^3}{e}^{\frac{-2r}{a_0}}\right)4\pi r^{2}dr}\\ \\ =\left(\frac{4}{a_0^3}\right){\displaystyle \underset{0}{\overset{2a_0}{\int }}{r}^2.e^{\frac{-2r}{a_0}}dr}\\ \\ =\left(\frac{4}{a_0^3}\right){\left[\frac{e^{\frac{-2r}{a_0}}}{\frac{-2}{a_0}}\left(r^2-\frac{2r}{\frac{-2}{a_0}}+\frac{2}{{\left(\frac{-2}{a_0}\right)}^2}\right)\right]}_0^{2a_0}\\ \\ =\left(\frac{4}{a_0^3}\right)\left[\frac{e^{\frac{-4a_0}{a_0}}}{\frac{-2}{a_0}}\left(4a_0^2-\frac{4a_0}{\frac{-2}{a_0}}+\frac{2}{{\left(\frac{-2}{a_0}\right)}^2}\right)-\left(\frac{1}{\frac{-2}{a_0}}\times \frac{2}{{\left(\frac{-2}{a_0}\right)}^2}\right)\right]\\ \\ =\left(\frac{4}{a_0^3}\right)\left[-\frac{a_0^{}{e}^{-4}}{2}\left(4a_0^2+2a_0^2+\frac{a_0^2}{2}\right)+\frac{a_0^3}{4}\right]\\ \\ =\left(\frac{4}{a_0^3}\right)\left[-3a_0^3{e}^{-4}+\frac{a_0^3{e}^{-4}}{4}+\frac{a_0^3}{4}\right]\\ \\ =\left(\frac{4}{a_0^3}\right)\left[\frac{a_0^3-13a_0^3{e}^{-4}}{4}\right]\\ \\ =1-5e^{-2}\\ \\ =0.76.\end{array}$

The probability is $0.76.$