Given Rydberg costant` R = 10^(5)cm^(-1)`. Supposing if electron jumps from M shell to K shell of H - atom, the frequency of the radiation emitted in cycle/s would be :-

Options

A. `(8)/(9)xx10^(5)`
B. `(8)/(3)xx10^(15)`
C. `(8)/(3)xx10^(11)`
D. `(8)/(9)xx10^(15)`

Explanation

To solve the problem of finding the frequency of radiation emitted when an electron jumps from the M shell to the K shell in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Shells and Their Principal Quantum Numbers:**
- The M shell corresponds to \( n = 3 \).
- The K shell corresponds to \( n = 1 \).

2. **Use the Rydberg Formula for Frequency:**
The formula for the frequency of radiation emitted during a transition between energy levels in a hydrogen atom is given by:
\[ \nu = R \cdot c \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \nu \) is the frequency,
- \( R \) is the Rydberg constant,
- \( c \) is the speed of light,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the principal quantum number of the lower energy level (K shell),
- \( n_2 \) is the principal quantum number of the higher energy level (M shell).

3. **Substitute the Values:**
- Given \( R = 10^5 \, \text{cm}^{-1} = 10^7 \, \text{m}^{-1} \) (conversion from cm to m),
- \( c = 3 \times 10^8 \, \text{m/s} \),
- \( Z = 1 \),
- \( n_1 = 1 \) (K shell),
- \( n_2 = 3 \) (M shell).

Now substituting these values into the formula:
\[ \nu = 10^7 \cdot 3 \times 10^8 \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]

4. **Calculate the Terms Inside the Parentheses:**
\[ \frac{1}{1^2} - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \]

5. **Combine and Simplify:**
\[ \nu = 10^7 \cdot 3 \times 10^8 \cdot \frac{8}{9} \]
\[ \nu = \frac{24 \times 10^{15}}{9} \]
\[ \nu = \frac{24}{9} \times 10^{15} = \frac{8}{3} \times 10^{15} \, \text{Hz} \]

6. **Final Result:**
The frequency of the radiation emitted when the electron jumps from the M shell to the K shell is:
\[ \nu = \frac{8}{3} \times 10^{15} \, \text{Hz} \]