`k_(a) ` wavelength emitted by an atom of atomic number `E= 11` is `lambda` find the atomic number for an atomic that amils `k_(a)` radiation with wavwlength `43`.

Options

A. `Z = 6`
B. `Z = 4`
C. `Z = 11`
D. `Z = 44`

Explanation

To solve the problem, we need to find the atomic number \( z_2 \) of an atom that emits K-alpha radiation with a wavelength of \( 4\lambda \), given that an atom with atomic number \( z_1 = 11 \) emits K-alpha radiation with a wavelength of \( \lambda \).

### Step-by-Step Solution:

1. **Understanding the relationship**: The wavelength \( \lambda \) of K-alpha radiation is related to the atomic number \( z \) by the formula:
\[ \lambda \propto \frac{1}{(z - 1)^2} \]
This implies:
\[ (z - 1)^2 \propto \frac{1}{\lambda} \]

2. **Setting up the equations**: For two different atomic numbers \( z_1 \) and \( z_2 \) with corresponding wavelengths \( \lambda_1 \) and \( \lambda_2 \):
\[ (z_1 - 1)^2 = K \cdot \lambda_1 \quad \text{and} \quad (z_2 - 1)^2 = K \cdot \lambda_2 \]
Here, \( K \) is a proportionality constant.

3. **Taking the ratio**: We can take the ratio of the two equations:
\[ \frac{(z_1 - 1)^2}{(z_2 - 1)^2} = \frac{\lambda_2}{\lambda_1} \]

4. **Substituting known values**: We know:
- \( z_1 = 11 \)
- \( \lambda_1 = \lambda \)
- \( \lambda_2 = 4\lambda \)

Substituting these values into the ratio gives:
\[ \frac{(11 - 1)^2}{(z_2 - 1)^2} = \frac{4\lambda}{\lambda} \]
Simplifying this, we have:
\[ \frac{(10)^2}{(z_2 - 1)^2} = 4 \]

5. **Calculating \( z_2 - 1 \)**: From the equation:
\[ \frac{100}{(z_2 - 1)^2} = 4 \]
We can rearrange this to find \( (z_2 - 1)^2 \):
\[ (z_2 - 1)^2 = \frac{100}{4} = 25 \]

6. **Solving for \( z_2 \)**: Taking the square root of both sides:
\[ z_2 - 1 = 5 \quad \text{or} \quad z_2 - 1 = -5 \]
Thus, we find:
\[ z_2 = 6 \quad \text{(since atomic number cannot be negative)} \]

### Final Answer:
The atomic number \( z_2 \) for the atom that emits K-alpha radiation with a wavelength of \( 4\lambda \) is \( 6 \).