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 will use the formula for the wavelength of K_alpha radiation emitted by an atom, which is given by:

\[ \lambda = \frac{1216 \, \text{Å}}{(Z - 1)^2} \]

where \(Z\) is the atomic number of the atom.

### Step 1: Calculate the wavelength for \(Z = 41\)

Given that the atomic number \(Z = 41\), we can substitute this value into the formula to find the wavelength \(\lambda\):

\[ \lambda = \frac{1216 \, \text{Å}}{(41 - 1)^2} = \frac{1216 \, \text{Å}}{40^2} = \frac{1216 \, \text{Å}}{1600} = \frac{1216}{1600} \, \text{Å} \]

### Step 2: Set up the equation for the new wavelength \(4\lambda\)

Now, we need to find the atomic number \(Z'\) for an atom that emits K_alpha radiation with a wavelength of \(4\lambda\). We can express this as:

\[ 4\lambda = \frac{1216 \, \text{Å}}{(Z' - 1)^2} \]

### Step 3: Substitute \(\lambda\) into the equation

Substituting the expression for \(\lambda\) from Step 1 into the equation for \(4\lambda\):

\[ 4 \left(\frac{1216}{1600}\right) = \frac{1216}{(Z' - 1)^2} \]

### Step 4: Simplify the equation

We can simplify this equation. First, multiply both sides by \((Z' - 1)^2\):

\[ 4 \left(\frac{1216}{1600}\right) (Z' - 1)^2 = 1216 \]

Now, divide both sides by \(1216\):

\[ 4 \left(\frac{1}{1600}\right) (Z' - 1)^2 = 1 \]

### Step 5: Solve for \(Z' - 1\)

Multiply both sides by \(1600\):

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

Now divide both sides by \(4\):

\[ (Z' - 1)^2 = 400 \]

Taking the square root of both sides gives:

\[ Z' - 1 = 20 \]

### Step 6: Find \(Z'\)

Adding \(1\) to both sides:

\[ Z' = 20 + 1 = 21 \]

### Conclusion

The atomic number \(Z'\) for the atom that emits K_alpha radiation with the wavelength \(4\lambda\) is:

\[ \boxed{21} \]