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)xx10^(5)`
B. `(3)/(16)xx10^(15)`
C. `(9)/(16)xx10^(15)`
D. `(3)/(4)xx10^(15)`

Answer

To find the frequency of the emitted radiation when an electron jumps from the 4th orbit to the 2nd orbit in a hydrogen atom, we can follow these steps:

### Step 1: Identify the given data
- Rydberg's constant, \( R = 10^5 \, \text{cm}^{-1} \)
- Initial orbit (n2) = 4
- Final orbit (n1) = 2

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

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

Substituting the values into the formula:

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

### Step 3: Calculate the terms inside the parentheses
Calculating \( \frac{1}{2^2} \) and \( \frac{1}{4^2} \):

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{4^2} = \frac{1}{16} = 0.0625 \]

Now, substituting these values:

\[ \frac{1}{\lambda} = 10^5 \left( 0.25 - 0.0625 \right) = 10^5 \left( 0.1875 \right) \]

### Step 4: Calculate \( \frac{1}{\lambda} \)
\[ \frac{1}{\lambda} = 10^5 \times 0.1875 = 18750 \, \text{cm}^{-1} \]

### Step 5: Find the wavelength \( \lambda \)
Taking the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{18750} \, \text{cm} \]

### Step 6: Convert wavelength to meters
To convert from centimeters to meters:

\[ \lambda = \frac{1}{18750} \times 10^{-2} \, \text{m} = \frac{10^{-2}}{18750} \, \text{m} \]

### Step 7: Calculate the frequency \( f \)
Using the speed of light \( c = 3 \times 10^8 \, \text{m/s} \), we can find the frequency using the formula:

\[ f = \frac{c}{\lambda} \]

Substituting the values:

\[ f = \frac{3 \times 10^8}{\frac{10^{-2}}{18750}} = 3 \times 10^8 \times \frac{18750}{10^{-2}} = 3 \times 10^8 \times 1875000 \]

### Step 8: Simplify the frequency calculation
Calculating \( 3 \times 1875000 \):

\[ 3 \times 1875000 = 5625000 \]

Thus,

\[ f = 5625000 \times 10^{10} \, \text{Hz} = 5.625 \times 10^{15} \, \text{Hz} \]

### Final Answer
The frequency of the emitted radiation is:

\[ f = \frac{9}{16} \times 10^{15} \, \text{Hz} \]