A H–atom in ground state has time period `T = 1.6 xx 10^(–16)` sec. find the frequency of electron in first excited state

Options

A. `7.8xx10^(14)`
B. `7.8xx10^(16)`
C. `3.7xx10^(14)`
D. `3.7xx10^(16)`

Answer

To find the frequency of the electron in the first excited state of 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 quantum number \( n \). For a hydrogen atom, the time period can be expressed as:
\[ T \propto n^3 \]
This means that the time period increases with the cube of the quantum number \( n \).

### Step 2: Write the relationship for the ground state and first excited state
Let \( T_1 \) be the time period for the ground state (\( n_1 = 1 \)) and \( T_2 \) be the time period for the first excited state (\( n_2 = 2 \)). We can express this relationship as:
\[ \frac{T_1}{T_2} = \left(\frac{n_1}{n_2}\right)^3 \]
Substituting \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{T_1}{T_2} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]
This implies:
\[ T_2 = 8 T_1 \]

### Step 3: Calculate the time period for the first excited state
Given that the time period for the ground state \( T_1 = 1.6 \times 10^{-16} \) seconds, we can find \( T_2 \):
\[ T_2 = 8 \times (1.6 \times 10^{-16}) = 12.8 \times 10^{-16} \text{ seconds} \]

### Step 4: Find the frequency
The frequency \( f \) is the reciprocal of the time period \( T \):
\[ f = \frac{1}{T} \]
Thus, the frequency for the first excited state \( f_2 \) is:
\[ f_2 = \frac{1}{T_2} = \frac{1}{12.8 \times 10^{-16}} \text{ Hz} \]

### Step 5: Calculate the frequency
Calculating \( f_2 \):
\[ f_2 = \frac{1}{12.8 \times 10^{-16}} \approx 7.8125 \times 10^{14} \text{ Hz} \]
Rounding this to two significant figures, we get:
\[ f_2 \approx 7.8 \times 10^{14} \text{ Hz} \]

### Final Answer
The frequency of the electron in the first excited state is approximately:
\[ \boxed{7.8 \times 10^{14} \text{ Hz}} \]