The ratio of the speed of electron in first Bohr orbit of H-atom to speed of light in vacuum is

Options

A. `137`
B. `7.30 xx 10^(-3)`
C. `100`
D. `10^(-2)`

Answer

To find the ratio of the speed of an electron in the first Bohr orbit of a hydrogen atom to the speed of light in vacuum, we can follow these steps:

### Step-by-step Solution:

1. **Determine the speed of the electron in the first Bohr orbit**:
According to the Bohr model, the speed \( v \) of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ v = \frac{2.188 \times 10^8 \, \text{cm/s}}{n^2} \]
For the first orbit, \( n = 1 \):
\[ v = \frac{2.188 \times 10^8 \, \text{cm/s}}{1^2} = 2.188 \times 10^8 \, \text{cm/s} \]

2. **Determine the speed of light in vacuum**:
The speed of light \( c \) is approximately:
\[ c = 3 \times 10^8 \, \text{m/s} \]
To convert this to centimeters per second:
\[ c = 3 \times 10^8 \, \text{m/s} \times 100 \, \text{cm/m} = 3 \times 10^{10} \, \text{cm/s} \]

3. **Calculate the ratio of the speed of the electron to the speed of light**:
Now, we can find the ratio \( R \) of the speed of the electron to the speed of light:
\[ R = \frac{v}{c} = \frac{2.188 \times 10^8 \, \text{cm/s}}{3 \times 10^{10} \, \text{cm/s}} \]
Simplifying this gives:
\[ R = \frac{2.188}{3} \times 10^{-2} = 0.7293 \times 10^{-2} = 7.293 \times 10^{-3} \]

4. **Final Result**:
Thus, the ratio of the speed of the electron in the first Bohr orbit of the hydrogen atom to the speed of light in vacuum is:
\[ R \approx 7.30 \times 10^{-3} \]

### Conclusion:
The final answer is approximately \( 7.30 \times 10^{-3} \).