An element X has atomic number 21. If a is constant the frequency of `K_alpha` line will be

Options

A. `200a^2`
B. `400a^2`
C. 200a
D. 400 a

Explanation

To find the frequency of the K_alpha line for the element X with atomic number 21, we can use the Moseley's law, which relates the frequency of the X-ray lines to the atomic number of the element.

### Step-by-Step Solution:

1. **Understanding Moseley's Law**:
Moseley's law states that the square root of the frequency (f) of the K_alpha line is proportional to the atomic number (Z) of the element, adjusted by a constant (A) and a screening constant (b). The formula can be expressed as:
\[ \sqrt{f} = A(Z - b) \]
For the K_alpha line, the screening constant \( b \) is equal to 1.

2. **Substituting Values**:
Given that the atomic number \( Z \) of element X is 21, we can substitute this value into the equation:
\[ \sqrt{f} = A(21 - 1) \]
Simplifying this gives:
\[ \sqrt{f} = A(20) \]

3. **Squaring Both Sides**:
To find the frequency \( f \), we square both sides of the equation:
\[ f = (A \cdot 20)^2 \]
This simplifies to:
\[ f = 400A^2 \]

4. **Conclusion**:
Therefore, the frequency of the K_alpha line for the element with atomic number 21 is:
\[ f = 400A^2 \]

### Final Answer:
The frequency of the K_alpha line will be \( 400A^2 \).

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