The atomic number (2) of an element whose k, wavelength is `lambda` is 11. The atomic number of an element whose wavelength is `4lambda` is equal to

Options

A. 6
B. 11
C. 44
D. 4

Answer

To solve the problem, we need to use the concept from Moseley's law, which relates the frequency of X-ray emissions to the atomic number of the element.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The atomic number \( Z_1 \) of the first element is 11.
- The wavelength \( \lambda_1 \) for this element is \( \lambda \).
- The wavelength \( \lambda_2 \) for the second element is \( 4\lambda \).

2. **Moseley's Law**:
- According to Moseley's law, the frequency \( \nu \) can be expressed as:
\[ \nu = A(Z - B)^2 \]
- For K-series, \( B = 1 \).

3. **Relate Wavelength and Frequency**:
- The frequency is related to the wavelength by:
\[ \nu = \frac{c}{\lambda} \]
- Therefore, we can write:
\[ \frac{c}{\lambda_1} = A(Z_1 - 1)^2 \]
\[ \frac{c}{\lambda_2} = A(Z_2 - 1)^2 \]

4. **Set Up the Ratio**:
- Taking the ratio of the two equations:
\[ \frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - 1)^2}{(Z_2 - 1)^2} \]

5. **Substituting Values**:
- Substitute \( \lambda_2 = 4\lambda \) and \( \lambda_1 = \lambda \):
\[ \frac{4\lambda}{\lambda} = \frac{(11 - 1)^2}{(Z_2 - 1)^2} \]
- This simplifies to:
\[ 4 = \frac{10^2}{(Z_2 - 1)^2} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 4(Z_2 - 1)^2 = 100 \]

7. **Dividing by 4**:
- Dividing both sides by 4:
\[ (Z_2 - 1)^2 = 25 \]

8. **Taking the Square Root**:
- Taking the square root of both sides:
\[ Z_2 - 1 = 5 \quad \text{or} \quad Z_2 - 1 = -5 \]
- Since atomic numbers cannot be negative, we take:
\[ Z_2 - 1 = 5 \]

9. **Solving for \( Z_2 \)**:
- Thus, we find:
\[ Z_2 = 5 + 1 = 6 \]

### Final Answer:
The atomic number of the element whose wavelength is \( 4\lambda \) is **6**.