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 will use Moseley's law, which relates the wavelength of X-rays emitted by an element to its atomic number. The law states that:

\[ \frac{1}{\lambda} \propto (Z - 1)^2 \]

Where:
- \(\lambda\) is the wavelength,
- \(Z\) is the atomic number.

### Step-by-Step Solution:

1. **Identify Given Values:**
- For the first element:
- Atomic number \(Z_1 = 11\)
- Wavelength \(\lambda_1 = \lambda\)
- For the second element:
- Wavelength \(\lambda_2 = 4\lambda\)
- Atomic number \(Z_2\) is what we need to find.

2. **Apply Moseley’s Law:**
- For the first element:
\[ \frac{1}{\lambda_1} = k(Z_1 - 1)^2 \]
Substituting the values:
\[ \frac{1}{\lambda} = k(11 - 1)^2 = k(10)^2 = 100k \]

- For the second element:
\[ \frac{1}{\lambda_2} = k(Z_2 - 1)^2 \]
Substituting the values:
\[ \frac{1}{4\lambda} = k(Z_2 - 1)^2 \]

3. **Relate the Two Equations:**
- From the first equation, we can express \(k\):
\[ k = \frac{1}{100\lambda} \]
- Substitute \(k\) into the second equation:
\[ \frac{1}{4\lambda} = \frac{1}{100\lambda}(Z_2 - 1)^2 \]

4. **Simplify the Equation:**
- Multiply both sides by \(100\lambda\):
\[ 25 = (Z_2 - 1)^2 \]

5. **Solve for \(Z_2\):**
- Taking the square root of both sides:
\[ Z_2 - 1 = 5 \quad \text{or} \quad Z_2 - 1 = -5 \]
- Thus:
\[ Z_2 = 6 \quad \text{(since atomic number cannot be negative)} \]

### Final Answer:
The atomic number of the element whose wavelength is \(4\lambda\) is \(Z_2 = 6\).

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