When an electron revolving in the ground state of a hydrogen atom jumps to its `2^"nd"` excited state.

Options

A. It frequency of revolution becomes `(3)^(2//3)` times
B. It frequency of revolution becomes `(3)^(-3)` times
C. Its speed becomes `(1/3)` times
D. Its speed becomes `(1/9)` times

Answer

To solve the problem of an electron in a hydrogen atom jumping from the ground state to its second excited state, we will follow these steps:

### Step 1: Identify the Initial and Final States
The ground state of a hydrogen atom corresponds to the principal quantum number \( n_1 = 1 \). The second excited state corresponds to \( n_2 = 3 \) (since the first excited state is \( n = 2 \)).

### Step 2: Understand the Relationship Between Frequency and Principal Quantum Number
The frequency of revolution of an electron in a hydrogen atom is inversely proportional to the cube of the principal quantum number. This can be expressed mathematically as:
\[ f \propto \frac{1}{n^3} \]
Thus, we can write the ratio of the frequencies for the two states as:
\[ \frac{f_1}{f_2} = \frac{n_2^3}{n_1^3} \]

### Step 3: Calculate the Frequencies
Substituting the values of \( n_1 \) and \( n_2 \):
\[ \frac{f_1}{f_2} = \frac{3^3}{1^3} = \frac{27}{1} = 27 \]
This implies:
\[ f_2 = \frac{f_1}{27} \]

### Step 4: Analyze the Speed of the Electron
The speed of the electron is inversely proportional to the principal quantum number:
\[ v \propto \frac{1}{n} \]
Thus, the ratio of the speeds can be expressed as:
\[ \frac{v_1}{v_2} = \frac{n_2}{n_1} \]
Substituting the values:
\[ \frac{v_1}{v_2} = \frac{3}{1} = 3 \]
This implies:
\[ v_2 = \frac{v_1}{3} \]

### Step 5: Conclusion
From the calculations:
- The frequency of revolution in the second excited state becomes \( \frac{1}{27} \) times that in the ground state.
- The speed of the electron in the second excited state becomes \( \frac{1}{3} \) times that in the ground state.

Thus, the correct options based on the question are:
- The frequency of revolution becomes \( \frac{1}{27} \) times.
- The speed becomes \( \frac{1}{3} \) times.