The frequency of revolution of electron in first excited state in Hydrogen is `7.8xx10^(14)s^(-1)`. The frequency of revolution of electron is ground state in `H^+` will be close to (in `s^(-1)` ):

Options

A. `6.2 xx 10^(15)`
B. `3.1 xx 10^(15)`
C. `1.2 xx 10^(16)`
D. `2.5 xx 10^(16)`

Explanation

To solve the problem, we need to determine the frequency of revolution of the electron in the ground state of \( H^+ \) using the information provided about the frequency in the first excited state of hydrogen.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the frequency of revolution of an electron in the first excited state (n=2) of hydrogen is given as \( f_1 = 7.8 \times 10^{14} \, \text{s}^{-1} \). We need to find the frequency of the electron in the ground state (n=1) of \( H^+ \).

2. **Using the Relationship Between Frequency and Principal Quantum Number**:
The frequency of revolution of an electron in a hydrogen-like atom is inversely proportional to the cube of the principal quantum number \( n \):
\[ f \propto \frac{1}{n^3} \]
Therefore, we can write:
\[ \frac{f_1}{f_2} = \frac{n_2^3}{n_1^3} \]
where \( f_1 \) is the frequency for \( n_1 = 2 \) (first excited state of hydrogen) and \( f_2 \) is the frequency for \( n_2 = 1 \) (ground state of \( H^+ \)).

3. **Substituting the Known Values**:
We know \( n_1 = 2 \) and \( n_2 = 1 \). Substituting these values into the equation gives:
\[ \frac{7.8 \times 10^{14}}{f_2} = \frac{1^3}{2^3} = \frac{1}{8} \]

4. **Cross-Multiplying to Solve for \( f_2 \)**:
Cross-multiplying gives:
\[ 7.8 \times 10^{14} = \frac{1}{8} f_2 \]
Therefore, we can rearrange to find \( f_2 \):
\[ f_2 = 8 \times (7.8 \times 10^{14}) \]

5. **Calculating \( f_2 \)**:
Now, calculating \( f_2 \):
\[ f_2 = 8 \times 7.8 \times 10^{14} = 62.4 \times 10^{14} = 6.24 \times 10^{15} \, \text{s}^{-1} \]

### Final Answer:
The frequency of revolution of the electron in the ground state of \( H^+ \) is approximately:
\[ f_2 \approx 6.24 \times 10^{15} \, \text{s}^{-1} \]