The time period of revolution of an electron in its ground state orbit in a hydrogen atom is `1.6 xx 10^(-16)` s. The frequency of the revoltuion in ( ` s^(-1)`). of the electron in its second exited state is

Options

A. `5.6xx10^(12)`
B. `7.8xx10^(16)`
C. `1.6xx10^(14)`
D. `2.3xx10^(14)`

Explanation

To find the frequency of revolution of the electron in its second excited state in a hydrogen atom, we can follow these steps:

### Step 1: Understand the relationship between time period and quantum number
The time period \( T \) of an electron in a hydrogen atom is related to the principal quantum number \( n \). The time period is proportional to \( n^3 \):
\[ T \propto n^3 \]

### Step 2: Set up the ratio of time periods
Let \( T_1 \) be the time period for the ground state (where \( n = 1 \)) and \( T_2 \) be the time period for the second excited state (where \( n = 3 \)). We can express this relationship as:
\[ \frac{T_1}{T_2} = \left(\frac{n_1}{n_2}\right)^3 \]
Substituting the values:
\[ \frac{T_1}{T_2} = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]

### Step 3: Calculate the time period for the second excited state
Given that the time period for the ground state \( T_1 = 1.6 \times 10^{-16} \) s, we can find \( T_2 \):
\[ T_2 = T_1 \times 27 = 1.6 \times 10^{-16} \times 27 \]
Calculating this gives:
\[ T_2 = 4.32 \times 10^{-15} \text{ s} \]

### Step 4: Find the frequency
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} \]
Thus, for the second excited state:
\[ f_2 = \frac{1}{T_2} = \frac{1}{4.32 \times 10^{-15}} \]
Calculating this gives:
\[ f_2 \approx 2.31 \times 10^{14} \text{ s}^{-1} \]

### Final Answer
The frequency of revolution of the electron in its second excited state is approximately:
\[ f_2 \approx 2.31 \times 10^{14} \text{ s}^{-1} \]

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