Chapter 28, Problem 37A

a.

To determine

To calculate: Radius of the orbital.

Answer to Problem 37A

Radius of the orbital = $\underset{\_}{4.77\times {10}^{-10}\text{ m}}$

Explanation of Solution

Given:

A hydrogen atom with an electron in the $n=3$ Bohr orbital.

Formula used:

The electron orbital radius in hydrogen is given by,

  $r=\frac{h^2{n}^2}{4{\pi }^{2}Km{q}^2}$

Where, $h$ is the Planck’s constant, $K$ is a constant from Coulomb’s Law and has a value of $9.0\times {10}^9{\text{ Nm}}^2/{\text{C}}^2$ , $m$ is the mass of electron and $q$ is the charge of an electron.

Calculation:

The electron orbital radius in hydrogen is calculated as:

  $\begin{array}{l}r=\frac{h^2{n}^2}{4{\pi }^{2}Km{q}^2}\\ \text{ =}\frac{{(6.626\times {10}^{-34}\text{ Js)}}^2\times {\left(3\right)}^2}{4{\pi }^2(9.0\times {10}^9{\text{ Nm}}^2/{\text{C}}^2)(9.11\times {10}^{-31}\text{ kg)(1}\text{.60}\times {\text{10}}^{-19}{\text{ C)}}^2}\\ \text{ = }4.77\times {10}^{-10}\text{ m}\end{array}$

Conclusion:

Radius of the orbital = $\underset{\_}{4.77\times {10}^{-10}\text{ m}}$

b.

To determine

To calculate: Electrostatic force between proton and electron.

Answer to Problem 37A

Electrostatic force between proton and electron = $\underset{\_}{1.01\times {10}^{-9}\text{ N}}$

Explanation of Solution

Given:

A hydrogen atom with an electron in the $n=3$ Bohr orbital.

Formula used:

Electrostatic force between proton and electron is given by,

  $F=\frac{K{q}^2}{r^2}$

Where $K$ is a constant from Coulomb’s Law and has a value of $9.0\times {10}^9{\text{ Nm}}^2/{\text{C}}^2$ , $q$ is the charge of an electron and $r$ is the orbital radius.

Calculation:

Electrostatic force between proton and electron is calculated as:

  $\begin{array}{l}F=\frac{K{q}^2}{r^2}\\ \text{ =}\frac{(9.0\times {10}^9{\text{ Nm}}^2/{\text{C}}^2){\text{(1}\text{.60}\times {\text{10}}^{-19}\text{ C)}}^2}{{(4.77\times {10}^{-10}\text{ m})}^2}\\ \text{ =}1.01\times {10}^{-9}\text{ N}\end{array}$

Conclusion:

Electrostatic force between proton and electron = $\underset{\_}{1.01\times {10}^{-9}\text{ N}}$

c.

To determine

To calculate: Centripetal acceleration of the electron.

Answer to Problem 37A

Centripetal acceleration of the electron = $\underset{\_}{\text{1}\text{.11}\times {\text{10}}^{21}{\text{ m/s}}^2}$

Explanation of Solution

Given:

A hydrogen atom with an electron in the $n=3$ Bohr orbital.

Formula used:

Acceleration is given by,

  $F=ma$

Where $F$ is force, $m$ is mass of an electron and $a$ is the acceleration.

Calculation:

Acceleration is calculated as:

  $\begin{array}{l}a=\frac{F}{m}\\ \text{ =}\frac{1.01\times {10}^{-9}\text{ N}}{9.11\times {10}^{-31}\text{ kg}}\\ \text{ =1}\text{.11}\times {\text{10}}^{21}{\text{ m/s}}^2\end{array}$

Conclusion:

Centripetal acceleration of the electron = $\underset{\_}{\text{1}\text{.11}\times {\text{10}}^{21}{\text{ m/s}}^2}$

d.

To determine

To calculate: Orbital electron speed and compare to speed of light.

Answer to Problem 37A

Orbital electron speed = $\underset{\_}{7.28\times {\text{10}}^5\text{ m/s}}$

The electron speed is $\underset{\_}{0.24\%\text{ of }c}$

Explanation of Solution

Given:

A hydrogen atom with an electron in the $n=3$ Bohr orbital.

Formula used:

Orbital electron speed is given by,

  $a=\frac{v^2}{r}$

Where $a$ is the acceleration, $v$ is the speed and $r$ is the orbital radius.

Calculation:

Orbital electron speed is calculated as:

  $\begin{array}{l}v=\sqrt{ar}\\ \text{ =}\sqrt{(1.11\times {10}^{21}{\text{ m/s}}^2)(4.77\times {10}^{-10}\text{ m)}}\\ \text{ =7}\text{.28}\times {\text{10}}^5\text{ m/s}\end{array}$

Comparing this with speed of light,

  $\frac{7.28\times {10}^5{\text{ m/s}}^2}{3\times {10}^8{\text{ m/s}}^2}\times 100=0.24\%\text{ of }c$

Conclusion:

Orbital electron speed = $\underset{\_}{7.28\times {\text{10}}^5\text{ m/s}}$

The electron speed is $\underset{\_}{0.24\%\text{ of }c}$