An electron in a hydrogen atom makes a transition `n_1 to n_2` where `n_1` and `n_2` are principle quantum numbers of the states . Assume the Bohr's model to be valid , the frequency of revolution in initial state is eight times that of final state. The ratio n`n_1/n_2` is

Options

A. `8:1`
B. `4:1`
C. `2:1`
D. `1:2`

Answer

To solve the problem, we need to find the ratio \( \frac{n_1}{n_2} \) given that the frequency of revolution in the initial state \( n_1 \) is eight times that of the final state \( n_2 \).

### Step-by-Step Solution:

1. **Understanding the Frequency Relation**:
According to Bohr's model, the frequency of revolution \( f \) of an electron in a hydrogen atom is given by:
\[ f \propto \frac{Z^2}{n^3} \]
where \( Z \) is the atomic number (which is 1 for hydrogen) and \( n \) is the principal quantum number.

2. **Setting Up the Frequencies**:
Let \( f_1 \) be the frequency of revolution in the initial state \( n_1 \) and \( f_2 \) be the frequency in the final state \( n_2 \). According to the problem:
\[ f_1 = 8 f_2 \]

3. **Expressing Frequencies in Terms of Quantum Numbers**:
From the frequency relation, we can express:
\[ f_1 = k \cdot \frac{1}{n_1^3} \quad \text{and} \quad f_2 = k \cdot \frac{1}{n_2^3} \]
where \( k \) is a constant.

4. **Substituting the Frequencies**:
Substituting the expressions for \( f_1 \) and \( f_2 \) into the equation \( f_1 = 8 f_2 \):
\[ k \cdot \frac{1}{n_1^3} = 8 \left( k \cdot \frac{1}{n_2^3} \right) \]

5. **Cancelling the Constant**:
Since \( k \) is common on both sides, we can cancel it out:
\[ \frac{1}{n_1^3} = 8 \cdot \frac{1}{n_2^3} \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ n_2^3 = 8 n_1^3 \]

7. **Taking the Cube Root**:
Taking the cube root of both sides:
\[ n_2 = 2 n_1 \]

8. **Finding the Ratio**:
Therefore, the ratio \( \frac{n_1}{n_2} \) is:
\[ \frac{n_1}{n_2} = \frac{n_1}{2 n_1} = \frac{1}{2} \]

### Final Answer:
The ratio \( \frac{n_1}{n_2} \) is \( \frac{1}{2} \).