The velocity of an electron in a certain Bohr orbit of H-atom bears 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 problem, we need to find the quantum number (n) of the orbit for an electron in a hydrogen atom, given that the velocity of the electron is in the ratio of 1:275 to the velocity of light.

### Step-by-Step Solution:

1. **Understand the given ratio**:
The velocity of the electron (v) is given as:
\[ \frac{v}{c} = \frac{1}{275} \]
where \(c\) is the speed of light.

2. **Calculate the velocity of the electron**:
The speed of light \(c\) is approximately \(3 \times 10^8 \, \text{m/s}\). Therefore, we can find the velocity of the electron:
\[ v = \frac{1}{275} \times c = \frac{1}{275} \times 3 \times 10^8 \, \text{m/s} \]
\[ v = \frac{3 \times 10^8}{275} \, \text{m/s} \]

3. **Use the formula for the velocity of an electron in a Bohr orbit**:
The velocity of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ v = \frac{2.16 \times 10^6 \times Z}{n} \, \text{m/s} \]
For hydrogen, \(Z = 1\), so the formula simplifies to:
\[ v = \frac{2.16 \times 10^6}{n} \, \text{m/s} \]

4. **Set the two expressions for velocity equal to each other**:
Now we equate the two expressions for velocity:
\[ \frac{3 \times 10^8}{275} = \frac{2.16 \times 10^6}{n} \]

5. **Rearrange to solve for n**:
Rearranging the equation gives:
\[ n = \frac{2.16 \times 10^6 \times 275}{3 \times 10^8} \]

6. **Calculate the value of n**:
Now we can compute the value of \(n\):
\[ n = \frac{2.16 \times 275}{3} \times 10^{-2} \]
\[ n = \frac{594}{3} \times 10^{-2} \]
\[ n = 198 \times 10^{-2} = 1.98 \approx 2 \]

7. **Conclusion**:
Therefore, the quantum number \(n\) of the orbit is:
\[ n = 2 \]

### Final Answer:
The quantum number \(n\) of the orbit is **2**.