If the wavelength of `K_(alpha)` radiation emitted an atom of atomic number Z = 41 is `lambda` , then the atomic number for an atom that emits `K_(alpha)` radiation with the wavelength `4lambda`, is

Options

A. 21
B. 28
C. 11
D. 36

Answer

To solve the problem, we need to establish the relationship between the wavelength of K-alpha radiation and the atomic number of the atom. The relationship can be expressed mathematically as:

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

Where \( \lambda \) is the wavelength and \( Z \) is the atomic number.

### Step 1: Establish the relationship for the first atom
For the first atom with atomic number \( Z_1 = 41 \) and wavelength \( \lambda_1 = \lambda \):

\[ \frac{1}{\lambda} \propto (Z_1 - 1)^2 \]

### Step 2: Establish the relationship for the second atom
For the second atom that emits K-alpha radiation with wavelength \( \lambda_2 = 4\lambda \):

\[ \frac{1}{4\lambda} \propto (Z_2 - 1)^2 \]

### Step 3: Set up the proportionality equations
From the proportionality, we can write:

\[ \frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - 1)^2}{(Z_2 - 1)^2} \]

Substituting the known values:

\[ \frac{4\lambda}{\lambda} = \frac{(41 - 1)^2}{(Z_2 - 1)^2} \]

This simplifies to:

\[ 4 = \frac{40^2}{(Z_2 - 1)^2} \]

### Step 4: Cross-multiply to solve for \( Z_2 \)
Cross-multiplying gives:

\[ 4(Z_2 - 1)^2 = 1600 \]

### Step 5: Divide both sides by 4
Dividing both sides by 4:

\[ (Z_2 - 1)^2 = 400 \]

### Step 6: Take the square root of both sides
Taking the square root of both sides:

\[ Z_2 - 1 = 20 \quad \text{or} \quad Z_2 - 1 = -20 \]

Since atomic numbers cannot be negative, we take:

\[ Z_2 - 1 = 20 \]

### Step 7: Solve for \( Z_2 \)
Adding 1 to both sides gives:

\[ Z_2 = 21 \]

### Final Answer
The atomic number for the atom that emits K-alpha radiation with the wavelength \( 4\lambda \) is \( Z_2 = 21 \).

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