Question

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately $\mathbf{2}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. Find (a) the force acting on the electron as it revolves in a circular orbit of radius $\mathbf{0}\mathbf{.}\mathbf{529}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathit{m}$and (b) the centripetal acceleration of the electron.

Short answer

The solution is

1. the force acting on the electron$8.33\times {10}^{-8}N$toward the nucleus.
2. the centripetal acceleration of the electron is$9.15\times {10}^{22}{\text{m/s}}^2$ inward.

Step-by-step solution

Given data

The speed of the electron is$v=2.20\times {10}^{6}m/s$

The radius of the orbit is $r=0.529\times {10}^{-10}m$

Concept Introduction

The centripetal force will act inward, causing centripetal acceleration.

The expression for the centripetal force is,

$\mathbf{\text{F =}}\frac{{\mathbf{\text{mv}}}^{\mathbf{\text{2}}}}{\mathbf{\text{r}}}$

Here, m is the mass of the electron, v is the velocity, r is the radius.

Calculation of the centripetal force

(a)

The centripetal force is,

$F=\frac{m{v}^2}{r}$

Substituterole="math" localid="1663755584561" $9.1\times {10}^{-31}kgform,2.20\times {10}^6\text{m/s}forv,and0.529\times {10}^{-10}mforr$

$$\begin{gathered} F=\frac{\left(9.1\times {10}^{-31}kg\right){\left(2.20\times {10}^6\text{m/s}\right)}^2}{0.529\times {10}^{-10}m} \\ =8.33\times {10}^{-8}N \end{gathered}$$

Hence, the force acting on the electron is $8.33\times {10}^{-8}N$toward the nucleus.

Calculation of the centripetal acceleration

(b)

The centripetal acceleration is,

$a=\frac{v^2}{r}$

Substitute $2.20\times {10}^{-6}\text{m/s}forvand0.529\times {10}^{-10}mforr.$

$$\begin{gathered} a=\frac{{\left(2.2\times {10}^6\text{m/s}\right)}^2}{0.53\times {10}^{10}m} \\ =9.15\times {10}^{22}{\text{m/s}}^2 \end{gathered}$$

Hence, the centripetal acceleration of the electron is $9.15\times {10}^{22}{\text{m/s}}^2$inward.