Chapter 27, Problem 63P

To determine

Radius and energy of the first Bohr orbit.

Answer to Problem 63P

Solution:

Radius of the first Bohr orbit will be $1.2\times {10}^{29}\text{m}$ and the radius of the first Bohr orbit will be $-4.2\times {10}^{-97}\text{J}$.

Explanation of Solution

Given:

Electron was bound to a proton by the gravitational force rather than by the electric force.

Formula Used:

1. $E=\frac{1}{2}{m}_e{v}^2-\frac{G{m}_e{m}_p}{r}$

2. $\frac{G{m}_e{m}_p}{r^2}=\frac{m_e{v}^2}{r}$

Where, $m_e$ is the mass of the electron.

$m_p$ is the mass of the proton.

G is Gravitational constant

r is the radius of the Bohr orbit

v is the speed of the electron.

h is Plank’s Constant

Calculation:

By using the formula $m_{e}vr=nh$, we can get the speed of the electron.

$\begin{array}{l}{m}_{e}vr=nh\\ r=\frac{nh}{m_{e}v}\end{array}$

On substituting the value of v in formula 2 $\frac{G{m}_e{m}_p}{r^2}=\frac{m_e{v}^2}{r}$, we get

$\frac{G{m}_e{m}_p}{r^2}=\frac{m_e}{r}\left(\frac{n^2{h}^2}{m_e^2{r}^2}\right)$

$\Rightarrow r=\left(\frac{h^2}{G{m}_e^2{m}_p}\right)n^2$

$\Rightarrow r=\left[\frac{{\left(6.6\times {10}^{-34}\text{J}\text{.s}\right)}^2}{\left(6.6\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right){\left(9.11\times {10}^{-31}\text{kg}\right)}^2\left(1.67\times {10}^{-27}\text{kg}\right)}\right](1^2)$

N=1 for hydrogen atom

So, $r=1.29\times {10}^{29}\text{m}$

Now for the energy of the first Bohr orbit

$\begin{array}{l}E=\frac{1}{2}{m}_e{v}^2-\frac{G{m}_e{m}_p}{r}=-\frac{1}{2}\left(\frac{G{m}_e{m}_p}{r}\right)\\ E_n=-\frac{1}{2}\left(G{m}_e{m}_p\right)\left(\frac{G{m}_e^2{m}_p}{h^2}\right)\frac{1}{n^2}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left(r=\left(\frac{h^2}{G{m}_e^2{m}_p}\right)n^2\right)\\ E_n=-\frac{1}{2}\frac{\left(G^2{m}_e^3{m}_p^2\right)}{h^2}\frac{1}{n^2}\\ E_1=-4.2\times {10}^{-97}\text{J}\end{array}$