`K_alpha` wavelength emitted by an atom of atomic number Z=11 is `lambda`. Find the atomic number for an atom that emits `K_alpha` radiation with wavelength `4lambda`.

Options

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

Explanation

To solve the problem, we will use Moseley's law which relates the wavelength of emitted X-rays to the atomic number of the element. Here’s a step-by-step solution:

### Step 1: Understand Moseley's Law
Moseley's law states that the frequency \( f \) of the emitted X-ray is proportional to the square of the atomic number \( Z \) minus a constant \( B \):
\[ f \propto (Z - B)^2 \]
For K-series X-rays, the constant \( B \) is typically taken as 1.

### Step 2: Relate Frequency and Wavelength
The frequency \( f \) is inversely related to the wavelength \( \lambda \):
\[ f = \frac{1}{\lambda} \]
Thus, we can rewrite Moseley's law in terms of wavelength:
\[ \frac{1}{\lambda} \propto (Z - 1)^2 \]
This implies:
\[ \lambda \propto \frac{1}{(Z - 1)^2} \]

### Step 3: Set Up the Equation for Given Wavelengths
Let \( \lambda_1 = \lambda \) for the atom with atomic number \( Z_1 = 11 \) and \( \lambda_2 = 4\lambda \) for the atom with atomic number \( Z_2 \). According to the proportionality established:
\[ \frac{\lambda_1}{\lambda_2} = \frac{(Z_2 - 1)^2}{(Z_1 - 1)^2} \]

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

### Step 5: Simplify and Solve for \( Z_2 \)
Cross-multiplying gives:
\[ (Z_2 - 1)^2 = \frac{100}{4} = 25 \]
Taking the square root of both sides:
\[ Z_2 - 1 = 5 \quad \text{or} \quad Z_2 - 1 = -5 \]
Thus, we have:
\[ Z_2 = 6 \quad \text{(valid solution)} \]
\[ Z_2 = -4 \quad \text{(not valid as atomic number cannot be negative)} \]

### Final Answer
The atomic number \( Z_2 \) for the atom that emits \( K_\alpha \) radiation with wavelength \( 4\lambda \) is:
\[ \boxed{6} \]