When in hydrogen like ion, electron jumps from n = 3, to n = 1, the emitted photon has frequency `2.7 xx 10^(15)Hz`. When electron jumps from n = 4 to n = 1, the frequency is

Options

A. `1.6 xx 10^(15)Hz`
B. `2.8 xx 10^(15)Hz`
C. `6.4 xx 10^(15)Hz`
D. `4.8 xx 10^(15)Hz`

Answer

To solve the problem, we will use the Rydberg formula for the frequency of emitted photons when an electron transitions between energy levels in a hydrogen-like ion. The formula is given by:

\[ f = R \cdot c \cdot \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

Where:
- \( f \) is the frequency of the emitted photon,
- \( R \) is the Rydberg constant,
- \( c \) is the speed of light,
- \( n_f \) is the final energy level,
- \( n_i \) is the initial energy level.

### Step 1: Identify the known values
From the problem, we know:
- For the transition from \( n_i = 3 \) to \( n_f = 1 \), the frequency \( f_1 = 2.7 \times 10^{15} \, \text{Hz} \).
- We need to find the frequency \( f_2 \) for the transition from \( n_i = 4 \) to \( n_f = 1 \).

### Step 2: Set up the proportional relationship
Since \( f \) is proportional to \( \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \), we can write:

\[ \frac{f_1}{f_2} = \frac{\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)_{n_i=3}}{\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)_{n_i=4}} \]

### Step 3: Calculate the values for \( n_i = 3 \) and \( n_i = 4 \)
For \( n_i = 3 \):
\[ \frac{1}{n_f^2} - \frac{1}{n_i^2} = \frac{1}{1^2} - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{8}{9} \]

For \( n_i = 4 \):
\[ \frac{1}{n_f^2} - \frac{1}{n_i^2} = \frac{1}{1^2} - \frac{1}{4^2} = 1 - \frac{1}{16} = \frac{15}{16} \]

### Step 4: Substitute the values into the proportional relationship
Now substituting the values into the proportional relationship:

\[ \frac{2.7 \times 10^{15}}{f_2} = \frac{\frac{8}{9}}{\frac{15}{16}} \]

### Step 5: Simplify the equation
Cross-multiplying gives:

\[ 2.7 \times 10^{15} \cdot \frac{15}{16} = f_2 \cdot \frac{8}{9} \]

Now, simplifying:

\[ f_2 = \frac{2.7 \times 10^{15} \cdot 15 \cdot 9}{16 \cdot 8} \]

### Step 6: Calculate \( f_2 \)
Calculating the right side:

\[ f_2 = \frac{2.7 \times 10^{15} \cdot 135}{128} \]

Calculating \( 2.7 \cdot 135 \):

\[ 2.7 \cdot 135 = 364.5 \]

Now, dividing by 128:

\[ f_2 = \frac{364.5 \times 10^{15}}{128} \approx 2.85 \times 10^{15} \, \text{Hz} \]

### Final Answer
Thus, the frequency when the electron jumps from \( n = 4 \) to \( n = 1 \) is approximately:

\[ f_2 \approx 2.85 \times 10^{15} \, \text{Hz} \]