The velocity of an electron in a certain Bohr orbit of H-atom beaes the ratio 1 : 275 to the velocity of light. The quantum number (n) of the orbit is

Options

A. 3
B. 2
C. 1
D. 4

Explanation

To solve the question regarding the velocity of an electron in a certain Bohr orbit of a hydrogen atom, we need to find the quantum number (n) given that the velocity of the electron bears the ratio 1:275 to the velocity of light (c).

### Step-by-Step Solution:

1. **Understanding the Velocity Ratio**:
Given that the velocity of the electron (v) is in the ratio of 1:275 to the speed of light (c), we can express this mathematically as:
\[ v = \frac{c}{275} \]

2. **Using the Bohr Model Formula**:
According to the Bohr model, the velocity of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ v = \frac{Z e^2}{2 \epsilon_0 h n} \]
For hydrogen (Z = 1), this simplifies to:
\[ v = \frac{e^2}{2 \epsilon_0 h n} \]

3. **Equating the Two Expressions for Velocity**:
We can set the two expressions for velocity equal to each other:
\[ \frac{e^2}{2 \epsilon_0 h n} = \frac{c}{275} \]

4. **Rearranging the Equation**:
Rearranging the equation to solve for n gives:
\[ n = \frac{275 e^2}{2 \epsilon_0 h c} \]

5. **Substituting Known Constants**:
We need to substitute the known values for the constants:
- \( e \) (electron charge) = \( 1.6 \times 10^{-19} \) C
- \( \epsilon_0 \) (permittivity of free space) = \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)

6. **Calculating n**:
Plugging in the values:
\[ n = \frac{275 \times (1.6 \times 10^{-19})^2}{2 \times (8.85 \times 10^{-12}) \times (6.626 \times 10^{-34}) \times (3 \times 10^8)} \]
After calculating, you will find that \( n \) is approximately equal to 1.

### Conclusion:
The quantum number (n) of the orbit is **1**.