An electron in a hydrogen atom makes a transition from `n_(1) to n_(2)`. If the time period of electron in the initial state is eight times that in the final state then Find the ratio `n_(1)/n_(2)`

Options

A. 2
B. 3
C. 4
D. 8

Answer

To solve the problem, we need to find the ratio \( \frac{n_1}{n_2} \) given that the time period of the electron in the initial state \( T_{n_1} \) is 8 times that in the final state \( T_{n_2} \).

### Step-by-Step Solution:

1. **Understanding the Time Period Relation**:
The time period \( T_n \) for an electron in a hydrogen atom can be expressed as:
\[ T_n \propto n^3 \]
This means that the time period is proportional to the cube of the principal quantum number \( n \).

2. **Setting Up the Ratio**:
Given that \( T_{n_1} = 8 T_{n_2} \), we can write:
\[ \frac{T_{n_1}}{T_{n_2}} = 8 \]

3. **Using the Proportionality**:
From the proportionality, we can express the ratio of time periods in terms of \( n_1 \) and \( n_2 \):
\[ \frac{T_{n_1}}{T_{n_2}} = \frac{n_1^3}{n_2^3} \]

4. **Equating the Two Ratios**:
Now we can set the two expressions for the ratio equal to each other:
\[ \frac{n_1^3}{n_2^3} = 8 \]

5. **Taking the Cube Root**:
Taking the cube root of both sides gives:
\[ \frac{n_1}{n_2} = \sqrt[3]{8} = 2 \]

6. **Final Result**:
Therefore, the ratio \( \frac{n_1}{n_2} \) is:
\[ \frac{n_1}{n_2} = 2 \]

### Conclusion:
The final answer is \( \frac{n_1}{n_2} = 2 \).