An electron in hydrogen atom jumps from `n_(1)` state to `n_(2)` state, where `n_(1)` and `n_(2)` represent the quantum number of two states. The time period of revolution of electron in initial state is 8 times that in final state. Then the ratio of `n_(1)" and "n_(2)` is

Options

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

Explanation

To solve the problem, we need to determine the ratio of the quantum numbers \( n_1 \) and \( n_2 \) based on the information given about the time periods of the electron's revolution in different states.

### Step-by-Step Solution:

1. **Understand the relationship between time period and quantum number**:
The time period \( T \) of an electron in a hydrogen atom is proportional to the cube of the principal quantum number \( n \). This can be expressed as:
\[ T \propto n^3 \]
Therefore, we can write:
\[ T_1 \propto n_1^3 \quad \text{and} \quad T_2 \propto n_2^3 \]

2. **Set up the ratio of the time periods**:
According to the problem, the time period of the electron in the initial state \( T_1 \) is 8 times that in the final state \( T_2 \):
\[ T_1 = 8 T_2 \]

3. **Express the ratio of time periods in terms of quantum numbers**:
Using the proportionality we established earlier, we can write:
\[ \frac{T_1}{T_2} = \frac{n_1^3}{n_2^3} \]
Substituting \( T_1 = 8 T_2 \) into the equation gives:
\[ 8 = \frac{n_1^3}{n_2^3} \]

4. **Rearranging the equation**:
Rearranging the equation gives:
\[ n_1^3 = 8 n_2^3 \]

5. **Taking the cube root**:
Taking the cube root of both sides results in:
\[ n_1 = 2 n_2 \]

6. **Finding the ratio of \( n_1 \) to \( n_2 \)**:
This can be expressed as:
\[ \frac{n_1}{n_2} = 2 \]
Therefore, the ratio \( n_1 : n_2 \) is:
\[ n_1 : n_2 = 2 : 1 \]

### Final Answer:
The ratio of \( n_1 \) to \( n_2 \) is \( 2 : 1 \).

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