An electron jumps from the `4th` orbit to the `2nd` orbit of hydrogen atom. Given the Rydberg's constant `R = 10^(5) cm^(-1)`. The frequency in `Hz` of the emitted radiation will be

Options

A. `(3)/(16) xx 10^(5)`
B. `(3)/(16) xx 10^(15)`
C. `(9)/(16) xx 10^(15)`
D. `(3)/(4) xx 10^(15)`

Explanation

To solve the problem of an electron jumping from the 4th orbit to the 2nd orbit of a hydrogen atom and finding the frequency of the emitted radiation, we can follow these steps:

### Step 1: Identify the values of n1 and n2
The initial orbit (n2) is 4 and the final orbit (n1) is 2.

### Step 2: Use the Rydberg formula
The Rydberg formula for the wavelength of emitted radiation is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting the values:
- \( R = 10^5 \, \text{cm}^{-1} \)
- \( n_1 = 2 \)
- \( n_2 = 4 \)

We get:

\[ \frac{1}{\lambda} = 10^5 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

### Step 3: Calculate the fractions
Calculating the fractions:

\[ \frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{4^2} = \frac{1}{16} \]

Now, substituting these values into the equation:

\[ \frac{1}{\lambda} = 10^5 \left( \frac{1}{4} - \frac{1}{16} \right) \]

### Step 4: Find a common denominator
The common denominator for 4 and 16 is 16:

\[ \frac{1}{4} = \frac{4}{16} \]

Thus,

\[ \frac{1}{\lambda} = 10^5 \left( \frac{4}{16} - \frac{1}{16} \right) = 10^5 \left( \frac{3}{16} \right) \]

### Step 5: Simplify the equation
This simplifies to:

\[ \frac{1}{\lambda} = \frac{3 \times 10^5}{16} \]

### Step 6: Calculate λ
Taking the reciprocal gives:

\[ \lambda = \frac{16}{3 \times 10^5} \, \text{cm} \]

### Step 7: Calculate the frequency
The frequency (ν) can be calculated using the speed of light (C):

\[ \nu = \frac{C}{\lambda} \]

Where \( C = 3 \times 10^{10} \, \text{cm/s} \).

Substituting the value of λ:

\[ \nu = \frac{3 \times 10^{10}}{\frac{16}{3 \times 10^5}} = 3 \times 10^{10} \times \frac{3 \times 10^5}{16} \]

### Step 8: Simplify the frequency
Calculating this gives:

\[ \nu = \frac{9 \times 10^{15}}{16} \approx 5.625 \times 10^{14} \, \text{Hz} \]

### Final Answer:
The frequency of the emitted radiation is approximately \( 5.625 \times 10^{14} \, \text{Hz} \).

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