Chapter 6, Problem 79PQ

The radius of circular electron orbits in the Bohr model of the hydrogen atom are given by (5.29 × 10^–11 m)n², where n is the electron's energy level (Fig. P6.79). The speed of the electron in each energy level is (c/137n), where c = 3 × 10⁸ m/s is the speed of light in vacuum, a. What is the centripetal acceleration of an electron in the ground state (n = 1) of the Bohr hydrogen atom? b. What are the magnitude and direction of the centripetal force acting on an electron in the ground state? c. What are the magnitude and direction of the centripetal force acting on an electron in the n = 2 excited state?

To determine

The centripetal acceleration of an electron in the ground state of the Bohr hydrogen atom.

Answer to Problem 79PQ

The centripetal acceleration of an electron in the ground state of the Bohr hydrogen atom is $9.06\times {10}^{22}{\text{m}/\text{s}}^2\text{inward}$ .

Explanation of Solution

Write the expression for the centripetal acceleration.

    $a_C=\frac{v^2}{r}$ (I)

Here, $a_C$ is the centripetal acceleration, $v$ is the speed of the electron, and $r$ is the radius of the orbit.

Use $c/137n$ for $v$ and $\left(5.29\times {10}^{-11}\text{m}\right)n^2$ for $r$ to rewrite (I).

    $a_C=\frac{{\left(c/137n\right)}^2}{\left(5.29\times {10}^{-11}\text{m}\right)n^2}$ (II)

Here, $c$ is the speed of light and $n$ is the electron energy level.

Conclusion:

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $1$ for $n$ to calculate $a_C$.

    $\begin{array}{c}{a}_C=\frac{{\left(3\times {10}^8\text{m}/\text{s}/137\left(1\right)\right)}^2}{\left(5.29\times {10}^{-11}\text{m}\right){\left(1\right)}^2}\\ =9.0645\times {10}^{22}{\text{m}/\text{s}}^2\\ \approx 9.06\times {10}^{22}{\text{m}/\text{s}}^2\end{array}$

Centripetal acceleration is directed towards the nucleus that is its acts inward.

Therefore, the centripetal acceleration of an electron in the ground state of the Bohr hydrogen atom is $9.06\times {10}^{22}{\text{m}/\text{s}}^2\text{inward}$ .

To determine

The magnitude and direction of the centripetal force acting on an electron in the in the ground state.

Answer to Problem 79PQ

The centripetal force acting on the electron in the ground state is $8.26\times {10}^{-8}\text{N}\text{inward}$ .

Explanation of Solution

Write the expression for the centripetal force.

    $F_C=m{a}_C$ (III)

Here, $F_C$ is the centripetal force and $m$ is the mass of the electron.

Use equation (II) in (III) to rewrite (III).

    $F_C=\frac{m{\left(c/137n\right)}^2}{\left(5.29\times {10}^{-11}\text{m}\right)n^2}$ (IV)

Conclusion:

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ , $9.11\times {10}^{-31}\text{kg}$ for $m$ and $1$ for $n$ to calculate $F_C$.

    $\begin{array}{c}{F}_C=\frac{9.11\times {10}^{-31}\text{kg}{\left(3\times {10}^8\text{m}/\text{s}/137\left(1\right)\right)}^2}{\left(5.29\times {10}^{-11}\text{m}\right){\left(1\right)}^2}\\ =8.2578\times {10}^{-8}\text{N}\\ \approx 8.26\times {10}^{-8}\text{N}\end{array}$

Centripetal force is directed towards the nucleus, that is its acts inward.

Therefore, the centripetal force acting on the electron in the ground state is $8.26\times {10}^{-8}\text{N}\text{inward}$ .

To determine

The magnitude and direction of the centripetal force acting on an electron in the in the ground state.

Answer to Problem 79PQ

The centripetal force acting on the electron in the $n=2$ excited state is $5.16\times {10}^{-9}\text{N}\text{inward}$ .

Explanation of Solution

Rewrite equation (IV).

    $F_C=\frac{m{\left(c/137n\right)}^2}{\left(5.29\times {10}^{-11}\text{m}\right)n^2}$ (IV)

For second excited state $n=2$.

Conclusion:

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ , $9.11\times {10}^{-31}\text{kg}$ for $m$ and $2$ for $n$ to calculate $F_C$.

    $\begin{array}{c}{F}_C=\frac{9.11\times {10}^{-31}\text{kg}{\left(3\times {10}^8\text{m}/\text{s}/137\left(2\right)\right)}^2}{\left(5.29\times {10}^{-11}\text{m}\right){\left(2\right)}^2}\\ =5.1611\times {10}^{-9}\text{N}\\ \approx 5.16\times {10}^{-9}\text{N}\end{array}$

Therefore, the centripetal force acting on the electron in the $n=2$ excited state is $5.16\times {10}^{-9}\text{N}\text{inward}$ .