Chapter 12, Problem 163CP

(a)

Interpretation Introduction

Interpretation:Relative probability to find electron in a sphere of volume $1.0\times {10}^{-3}{\text{ pm}}^3$ centered at nucleus is to be calculated.

Concept introduction:Wavefunction is defined as the function of wave that gives information about the properties of wave. The expression for wavefunction of $1s$ orbital for a sphere is as follows:

  ${\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }}{\left(\frac{Z}{a_0}\right)}^{\frac{3}{2}}\text{exp}\left(\frac{-r}{a_0}\right)$

Square of wavefunction of sphere gives the probability of finding electron. Probability is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Answer to Problem 163CP

Probability to findelectron in nucleus is $2.2\times {10}^{-9}$ .

Explanation of Solution

Square of wavefunction of sphere gives the probability of finding electron. Probability at nucleus is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Where,

• ${\psi }_{1s}$ iswavefunction of $1s$ orbital.
• $Z$is atomic number of hydrogen atom.
• $a_0$is Bohr’s radius.
• $r$is distance from nucleus.
• $\pi$is constant term.

Value of $Z$ is 1.

Value of $a_0$ is $0.0529\text{ nm}$ .

Value of $r$ is 0.

Value of $\pi$ is 3.14.

Substitute values in above equation.

  $\begin{array}{c}{\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)\\ =\frac{1}{3.14}{\left(\frac{1}{0.0529\text{ nm}}\right)}^3\text{exp}\left(\frac{-2\left(0\right)}{0.0529\text{ nm}}\right)\\ =2150.49{\left(\text{nm}\right)}^{-3}{\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)}^3\\ =2.15\times {10}^{30}{\text{ m}}^{-3}\end{array}$

Total probability is the product of probability of electron at nucleus and volume of sphere. It is expressed as follows:

  $\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)$

Where,

• ${\psi }_{1s}^2$is probability of finding electron at nucleus.
• $V$is volume of sphere.

Value of ${\psi }_{1s}^2$ is $2.15\times {10}^{30}{\text{ m}}^{-3}$ .

Value of $V$ is $1.0\times {10}^{-3}{\text{ pm}}^3$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)\\ =\left(2.15\times {10}^{30}{\text{ m}}^{-3}\right)\left(1.0\times {10}^{-3}{\text{ pm}}^3\right)\left(\frac{{10}^{-36}{\text{ m}}^3}{1{\text{ pm}}^3}\right)\\ =2.2\times {10}^{-9}\end{array}$

Hence, probability of finding electron in nucleus is $2.2\times {10}^{-9}$ .

(b)

Interpretation Introduction

Interpretation:Relative probability to findelectron in a sphere of volume $1.0\times {10}^{-3}{\text{ pm}}^3$ centered on a point $1.0\times {10}^{-11}\text{ m}$ from nucleus is to be calculated.

Concept introduction:Wavefunction is defined as the function of wave that gives information about the properties of wave. The expression for wavefunction of $1s$ orbital for a sphere is as follows:

  ${\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }}{\left(\frac{Z}{a_0}\right)}^{\frac{3}{2}}\text{exp}\left(\frac{-r}{a_0}\right)$

Square of wavefunction of sphere gives the probability of finding electron. Probability is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Answer to Problem 163CP

Probability to findelectron at $1.0\times {10}^{-11}\text{ m}$ from nucleus is $1.5\times {10}^{-9}$ .

Explanation of Solution

Square of wavefunction of sphere gives the probability of finding electron. Probability of electron centered on a point $1.0\times {10}^{-11}\text{ m}$ from nucleus is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Where,

• ${\psi }_{1s}$ iswavefunction of $1s$ orbital.
• $Z$is atomic number of hydrogen atom.
• $a_0$is Bohr’s radius.
• $r$is distance from nucleus.
• $\pi$is constant term.

Value of $Z$ is 1.

Value of $a_0$ is $0.0529\text{ nm}$ .

Value of $r$ is $1.0\times {10}^{-11}\text{ m}$ .

Value of $\pi$ is 3.14.

Substitute values in above equation.

  $\begin{array}{c}{\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)\\ =\frac{1}{3.14}{\left(\frac{1}{0.0529\text{ nm}}\right)}^3\text{exp}\left(\frac{-2\left(1.0\times {10}^{-11}\text{ m}\right)}{0.0529\text{ nm}}\right){\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)}^3\\ =1.5\times {10}^{30}{\text{ m}}^{-3}\end{array}$

Total probability is the product of probability of electron at nucleus and volume of sphere. It is expressed as follows:

  $\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)$

Where,

• ${\psi }_{1s}^2$ is probability of finding electron at $1.0\times {10}^{-11}\text{ m}$ from nucleus.
• $V$is volume of sphere.

Value of ${\psi }_{1s}^2$ is $1.5\times {10}^{30}{\text{ m}}^{-3}$ .

Value of $V$ is $1.0\times {10}^{-3}{\text{ pm}}^3$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)\\ =\left(1.5\times {10}^{30}{\text{ m}}^{-3}\right)\left(1.0\times {10}^{-3}{\text{ pm}}^3\right)\left(\frac{{10}^{-36}{\text{ m}}^3}{1{\text{ pm}}^3}\right)\\ =1.5\times {10}^{-9}\end{array}$

Hence, probability to findelectron at $1.0\times {10}^{-11}\text{ m}$ from nucleusis $1.5\times {10}^{-9}$ .

(c)

Interpretation Introduction

Interpretation:Relative probability to findelectron in a sphere of volume $1.0\times {10}^{-3}{\text{ pm}}^3$ centered on a point $53\text{ pm}$ from nucleus is to be calculated.

Concept introduction:Wavefunction is defined as the function of wave that gives information about the properties of wave. The expression for wavefunction of $1s$ orbital for a sphere is as follows:

  ${\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }}{\left(\frac{Z}{a_0}\right)}^{\frac{3}{2}}\text{exp}\left(\frac{-r}{a_0}\right)$

Square of wavefunction of sphere gives the probability of finding electron. Probability is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Answer to Problem 163CP

Probability to findelectron on a point $53\text{ pm}$ from nucleus is $2.9\times {10}^{-10}$ .

Explanation of Solution

Square of wavefunction of sphere gives the probability of finding electron. Probability of electron centered on a point $53\text{ pm}$ from nucleus is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Where,

• ${\psi }_{1s}$ iswavefunction of $1s$ orbital.
• $Z$is atomic number of hydrogen atom.
• $a_0$is Bohr’s radius.
• $r$is distance from nucleus.
• $\pi$is constant term.

Value of $Z$ is 1.

Value of $a_0$ is $0.0529\text{ nm}$ .

Value of $r$ is $53\text{ pm}$ .

Value of $\pi$ is 3.14.

Substitute values in above equation.

  $\begin{array}{c}{\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)\\ =\frac{1}{3.14}{\left(\frac{1}{0.0529\text{ nm}}\right)}^3\text{exp}\left(\frac{-2\left(53\text{ pm}\right)\left(\frac{{10}^{-12}\text{ m}}{1\text{ pm}}\right)}{0.0529\text{ nm}}\right){\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)}^3\\ =2.9\times {10}^{29}{\text{ m}}^{-3}\end{array}$

Total probability is the product of probability of electron at nucleus and volume of sphere. It is expressed as follows:

  $\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)$

Where,

• ${\psi }_{1s}^2$ is probability of finding electron at $53\text{ pm}$ from nucleus.
• $V$is volume of sphere.

Value of ${\psi }_{1s}^2$ is $2.9\times {10}^{29}{\text{ m}}^{-3}$ .

Value of $V$ is $1.0\times {10}^{-3}{\text{ pm}}^3$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)\\ =\left(2.9\times {10}^{29}{\text{ m}}^{-3}\right)\left(1.0\times {10}^{-3}{\text{ pm}}^3\right)\left(\frac{{10}^{-36}{\text{ m}}^3}{1{\text{ pm}}^3}\right)\\ =2.9\times {10}^{-10}\end{array}$

Hence, probability of finding electron at $53\text{ pm}$ from nucleus is $2.9\times {10}^{-10}$ .

(d)

Interpretation Introduction

Interpretation:Relative probability to find electron in two concentric spheres of radius $10.05\text{ pm}$ and $9.95\text{ pm}$ is to be calculated.

Concept introduction:Wavefunction is defined as the function of wave that gives information about the properties of wave. The expression for wavefunction of $1s$ orbital for a sphere is as follows:

  ${\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }}{\left(\frac{Z}{a_0}\right)}^{\frac{3}{2}}\text{exp}\left(\frac{-r}{a_0}\right)$

Square of wavefunction of sphere gives the probability of finding electron. Probability is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Answer to Problem 163CP

Probability to findelectronin two concentric spheres of radius $10.05\text{ pm}$ and $9.95\text{ pm}$ is $1.9\times {10}^{-4}$ .

Explanation of Solution

Volume of sphere from two concentric spheres is calculated by subtraction of volume of both spheres. Volume is expressed as follows:

  $\begin{array}{c}\text{Volume}=\frac{4}{3}\pi r_1^3-\frac{4}{3}\pi r_2^3\\ =\frac{4}{3}\pi \left(r_1^3-r_2^3\right)\end{array}$

Where,

• $\pi$is constant term.
• $r_1$is radius of one sphere.
• $r_2$is radius of second sphere.

Value of $\pi$ is 3.14.

Value of $r_1$ is $10.05\text{ pm}$ .

Value of $r_2$ is $9.95\text{ pm}$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Volume}=\frac{4}{3}\pi \left(r_1^3-r_2^3\right)\\ =\frac{4}{3}\left(3.14\right)\left({\left(10.05\text{ pm}\right)}^3-{\left(9.95\text{ pm}\right)}^3\right){\left(\frac{{10}^{-12}\text{ m}}{\text{ 1 pm}}\right)}^3\\ =1.3\times {10}^{-34}{\text{ m}}^3\end{array}$

Square of wavefunction of sphere gives the probability of finding electron at center of nucleus. Probability is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Where,

• ${\psi }_{1s}$ iswavefunction of $1s$ orbital.
• $Z$is atomic number of hydrogen atom.
• $a_0$is Bohr’s radius.
• $r$is distance from nucleus.
• $\pi$is constant term.

Value of $Z$ is 1.

Value of $a_0$ is $0.0529\text{ nm}$ .

Value of $r$ is $10.05\text{ pm}$ .

Value of $\pi$ is 3.14.

Substitute values in above equation.

  $\begin{array}{c}{\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)\\ =\frac{1}{3.14}{\left(\frac{1}{0.0529\text{ nm}}\right)}^3\text{exp}\left(\frac{-2\left(10.05\text{ pm}\right)\left(\frac{{10}^{-12}\text{ m}}{1\text{ pm}}\right)}{0.0529\text{ nm}}\right)\left(\frac{{10}^{-36}{\text{ m}}^3}{1{\text{ pm}}^3}\right)\\ =1.47\times {10}^{30}{\text{ m}}^{-3}\end{array}$

Total probability is the product of probability of electron at nucleus and volume of sphere. It is expressed as follows:

  $\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)$

Where,

• ${\psi }_{1s}^2$is probability of finding electron.
• $V$is volume of sphere.

Value of ${\psi }_{1s}^2$ is $1.47\times {10}^{30}{\text{ m}}^{-3}$ .

Value of $V$ is $1.3\times {10}^{-34}{\text{ m}}^3$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)\\ =\left(1.47\times {10}^{30}{\text{ m}}^{-3}\right)\left(1.3\times {10}^{-34}{\text{ m}}^3\right)\\ =1.9\times {10}^{-4}\end{array}$

Hence, probability to findelectron is $1.9\times {10}^{-4}$ .

(e)

Interpretation Introduction

Interpretation:Relative probability to find electron in two concentric spheres of radius $52.85\text{ pm}$ and $52.95\text{ pm}$ is to be calculated.

Concept introduction:Wavefunction is defined as the function of wave that gives information about the properties of wave. The expression for wavefunction of $1s$ orbital for a sphere is as follows:

  ${\psi }_{\text{1s}}=\frac{1}{\sqrt{\pi }}{\left(\frac{Z}{a_0}\right)}^{\frac{3}{2}}\text{exp}\left(\frac{-r}{a_0}\right)$

Square of wavefunction of sphere gives the probability of finding electron. Probability is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Answer to Problem 163CP

Probability to findelectron in two concentric spheres of radius $52.85\text{ pm}$ and $52.95\text{ pm}$ is $1.0\times {10}^{-3}$ .

Explanation of Solution

Volume of sphere from two concentric spheres is calculated by subtraction of volume of both spheres. Volume is expressed as follows:

  $\begin{array}{c}\text{Volume}=\frac{4}{3}\pi r_1^3-\frac{4}{3}\pi r_2^3\\ =\frac{4}{3}\pi \left(r_1^3-r_2^3\right)\end{array}$

Where,

• $\pi$is constant term.
• $r_1$is radius of one sphere.
• $r_2$is radius of second sphere.

Value of $\pi$ is 3.14.

Value of $r_1$ is $52.95\text{ pm}$ .

Value of $r_2$ is $52.85\text{ pm}$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Volume}=\frac{4}{3}\pi \left(r_1^3-r_2^3\right)\\ =\frac{4}{3}\left(3.14\right)\left({\left(52.95\text{ pm}\right)}^3-{\left(52.85\text{ pm}\right)}^3\right){\left(\frac{{10}^{-12}\text{ m}}{\text{ 1 pm}}\right)}^3\\ =4\times {10}^{-33}{\text{ m}}^3\end{array}$

Square of wavefunction of sphere gives the probability of finding electron at center of nucleus. Probability is expressed as follows:

  ${\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)$

Where,

• ${\psi }_{1s}$ iswavefunction of $1s$ orbital.
• $Z$is atomic number of hydrogen atom.
• $a_0$is Bohr’s radius.
• $r$is distance from nucleus.
• $\pi$is constant term.

Value of $Z$ is 1.

Value of $a_0$ is $0.0529\text{ nm}$ .

Value of $r$ is $52.95\text{ pm}$ .

Value of $\pi$ is 3.14.

Substitute values in above equation.

  $\begin{array}{c}{\psi }_{1s}^2=\frac{1}{\pi }{\left(\frac{Z}{a_0}\right)}^3\text{exp}\left(\frac{-2r}{a_0}\right)\\ =\frac{1}{3.14}{\left(\frac{1}{0.0529\text{ nm}}\right)}^3\text{exp}\left(\frac{-2\left(52.95\text{ pm}\right)\left(\frac{{10}^{-12}\text{ m}}{1\text{ pm}}\right)}{0.0529\text{ nm}}\right)\left(\frac{{10}^{-36}{\text{ m}}^3}{1{\text{ pm}}^3}\right)\\ =2.91\times {10}^{29}{\text{ m}}^{-3}\end{array}$

Total probability is the product of probability of electron at nucleus and volume of sphere. It is expressed as follows:

  $\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)$

Where,

• ${\psi }_{1s}^2$is probability of finding electron.
• $V$is volume of sphere.

Value of ${\psi }_{1s}^2$ is $2.91\times {10}^{29}{\text{ m}}^{-3}$ .

Value of $V$ is $4\times {10}^{-33}{\text{ m}}^3$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Total probability}=\left({\psi }_{1s}^2\right)\left(V\right)\\ =\left(1.47\times {10}^{30}{\text{ m}}^{-3}\right)\left(4\times {10}^{-33}{\text{ m}}^3\right)\\ =1.0\times {10}^{-3}\end{array}$

Hence, probability of finding electron is $1.0\times {10}^{-3}$ .