The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is

Options

A. `1//2`
B. `2//137`
C. `1//137`
D. `1//237`

Explanation

To find the ratio of the speed of the electron in the ground state of hydrogen to the speed of light in vacuum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the speed of the electron in the nth orbit of hydrogen**:
The speed of the electron in the nth orbit of a hydrogen atom is given by the formula:
\[ V_n = \frac{e^2}{2 \epsilon_0 n h} \]
where:
- \( e \) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \( \epsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)),
- \( h \) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \( n \) is the principal quantum number (for ground state, \( n = 1 \)).

2. **Substituting the values for the ground state (n=1)**:
For the ground state, we substitute \( n = 1 \) into the formula:
\[ V_1 = \frac{e^2}{2 \epsilon_0 h} \]

3. **Calculate \( V_1 \)**:
Substitute the known values:
\[ V_1 = \frac{(1.6 \times 10^{-19})^2}{2 \times (8.85 \times 10^{-12}) \times (6.63 \times 10^{-34})} \]

4. **Calculate the speed of light \( c \)**:
The speed of light in vacuum is:
\[ c = 3 \times 10^8 \, \text{m/s} \]

5. **Calculate the ratio \( \frac{V_1}{c} \)**:
Now, we find the ratio of the speed of the electron in the ground state to the speed of light:
\[ \frac{V_1}{c} = \frac{V_1}{3 \times 10^8} \]

6. **Final calculation**:
After performing the calculations, we find that:
\[ \frac{V_1}{c} = \frac{1}{37} \]

### Conclusion:
The ratio of the speed of the electron in the ground state of hydrogen to the speed of light in vacuum is:
\[ \frac{V_1}{c} = \frac{1}{37} \]