The period of revolution of electron in the third orbit in a H-atom is `4.132 xx 10^(-15)s`. Hence the period in the fourth orbit is

Options

A. `9.794 xx 10^(-15)s`
B. `9.794 xx 10^(-14)s`
C. `9.974 xx 10^(-15)s`
D. `9.974 xx 10^(-14)s`

Explanation

To find the period of revolution of the electron in the fourth orbit of a hydrogen atom, we can use the relationship between the period of revolution and the principal quantum number (n). The period of revolution (T) is directly proportional to the cube of the principal quantum number (n³).

### Step-by-Step Solution:

1. **Identify the given data:**
- The period of revolution of the electron in the third orbit (n=3) is given as:
\[ T_3 = 4.132 \times 10^{-15} \, \text{s} \]

2. **Use the proportionality relation:**
- The period of revolution is proportional to \( n^3 \):
\[ T_n \propto n^3 \]
- Therefore, we can express the periods for different orbits as:
\[ \frac{T_4}{T_3} = \frac{n_4^3}{n_3^3} \]

3. **Substituting the values of n:**
- For the third orbit, \( n_3 = 3 \)
- For the fourth orbit, \( n_4 = 4 \)
- Substitute these values into the equation:
\[ \frac{T_4}{T_3} = \frac{4^3}{3^3} = \frac{64}{27} \]

4. **Rearranging to find \( T_4 \):**
- Now, we can express \( T_4 \) in terms of \( T_3 \):
\[ T_4 = T_3 \cdot \frac{64}{27} \]

5. **Substituting the value of \( T_3 \):**
- Substitute \( T_3 = 4.132 \times 10^{-15} \, \text{s} \):
\[ T_4 = 4.132 \times 10^{-15} \cdot \frac{64}{27} \]

6. **Calculating \( T_4 \):**
- First, calculate \( \frac{64}{27} \):
\[ \frac{64}{27} \approx 2.37037 \]
- Now multiply:
\[ T_4 \approx 4.132 \times 10^{-15} \times 2.37037 \approx 9.794 \times 10^{-15} \, \text{s} \]

7. **Final answer:**
- The period of revolution of the electron in the fourth orbit is:
\[ T_4 \approx 9.794 \times 10^{-15} \, \text{s} \]