Question

The ratio of the speed of the electron in the ground state of a hydrogen atom to the speed of light is

Answer 1

Speed of the electron in the ground state.

The speed of the electron in the ground state is the numeric value of the velocity of the electron in the 1st orbit. It is equal to $2.19\times {10}^{6}m/s$

Step 1: Given:

The electron is in the ground state, so $n=1$

Step 2: Formula Used:

The speed of revolving electron in n^th state of hydrogen atom = $v=\frac{e^2}{2nh{\in }_0}$

Here $e$ is charge on electron, $n$ is state of electron, $h$ is plank's constant, ${\in }_0$ is permeability free space.

Step 3: Calculating the speed of the electron in hydrogen ground state:

For ground state, $n=1$

$$\begin{gathered} v=\frac{{(1.6\times {10}^{-19})}^2}{2\left(1\right)(6.6\times {10}^{-34})(8.85\times {10}^{-12})} \\ v=\frac{2.56\times {10}^{-38}}{116.82\times {10}^{-46}} \\ v=0.0219\times {10}^{8}m/sor2.19\times {10}^{6}m/s \end{gathered}$$

Step 4: Calculating the ratio of the speed of the electron in hydrogen ground state to the speed of light:

Speed of light, $c=3\times {10}^8$

Hence, The ratio of the speed of the electron in the ground state of a hydrogen atom to the speed of light is

$$\begin{gathered} \frac{v}{c}=\frac{0.0219\times {10}^8}{3\times {10}^8} \\ \frac{v}{c}=\frac{1}{137} \end{gathered}$$

Final Answer: The ratio of the speed of the electron in the ground state of a hydrogen atom to the speed of light is $\frac{1}{137}$