The ratio of wavelength of `K_alpha` line of an element with atomic number 61 to wavelength of `K_alpha` line of an element with atomic number 21 is

Options

A. `3:1`
B. `4:1`
C. `1:9`
D. `2:1`

Explanation

To find the ratio of the wavelengths of the K-alpha line of an element with atomic number 61 to that of an element with atomic number 21, we can use the modified formula based on Moseley's theory.

### Step-by-Step Solution:

1. **Understanding K-alpha Transition**:
The K-alpha line corresponds to an electron transition from the n=2 energy level to the n=1 energy level in an atom.

2. **Using the Formula**:
According to Moseley's modification of the Bohr model, the wavelength (λ) of the emitted radiation can be expressed as:
\[ \frac{1}{\lambda} = k \cdot (Z - 1)^2 \cdot \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where:
- \( Z \) is the atomic number,
- \( n_f \) is the final energy level (1 for K-alpha),
- \( n_i \) is the initial energy level (2 for K-alpha),
- \( k \) is a constant.

3. **Setting Up the Equations**:
For the element with atomic number 61:
\[ \frac{1}{\lambda_1} = k \cdot (61 - 1)^2 \cdot \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Simplifying this gives:
\[ \frac{1}{\lambda_1} = k \cdot (60)^2 \cdot \left( 1 - \frac{1}{4} \right) = k \cdot 3600 \cdot \frac{3}{4} = k \cdot 2700 \]

For the element with atomic number 21:
\[ \frac{1}{\lambda_2} = k \cdot (21 - 1)^2 \cdot \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Simplifying this gives:
\[ \frac{1}{\lambda_2} = k \cdot (20)^2 \cdot \left( 1 - \frac{1}{4} \right) = k \cdot 400 \cdot \frac{3}{4} = k \cdot 300 \]

4. **Finding the Ratio of Wavelengths**:
Now, we can find the ratio of the wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{\lambda_2}}{\frac{1}{\lambda_1}} = \frac{k \cdot 300}{k \cdot 2700} \]
The \( k \) cancels out:
\[ \frac{\lambda_1}{\lambda_2} = \frac{300}{2700} = \frac{1}{9} \]

5. **Final Ratio**:
Thus, the ratio of the wavelengths of the K-alpha line of the element with atomic number 61 to that of the element with atomic number 21 is:
\[ \lambda_1 : \lambda_2 = 1 : 9 \]