if the frequency fo `K_a` X-ray emitted from the element with atomic number 31 is f, then the frequency of `K_a` x-ray emitted from the elemet with atomic number 51 would be

Options

A. `(5f)/(3)`
B. `(51f)/(31)`
C. `(9f)/(25)`
D. `(25f)/(9)`

Explanation

To solve the problem, we will use Moseley's law which relates the frequency of characteristic 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 difference between the atomic number \( Z \) and a constant \( b \):
\[ f \propto (Z - b)^2 \]
For K-series X-rays, the value of \( b \) is typically taken as 1.

### Step 2: Write the equation for the first element
For the element with atomic number \( Z_1 = 31 \):
\[ f = k \cdot (Z_1 - b)^2 \]
Substituting \( b = 1 \):
\[ f = k \cdot (31 - 1)^2 = k \cdot (30)^2 \]

### Step 3: Write the equation for the second element
For the element with atomic number \( Z_2 = 51 \):
\[ f' = k \cdot (Z_2 - b)^2 \]
Substituting \( b = 1 \):
\[ f' = k \cdot (51 - 1)^2 = k \cdot (50)^2 \]

### Step 4: Set up the ratio of frequencies
Now we can set up the ratio of the frequencies \( f' \) and \( f \):
\[ \frac{f'}{f} = \frac{k \cdot (50)^2}{k \cdot (30)^2} \]
The \( k \) cancels out:
\[ \frac{f'}{f} = \frac{(50)^2}{(30)^2} = \frac{2500}{900} = \frac{25}{9} \]

### Step 5: Solve for \( f' \)
Now, we can express \( f' \) in terms of \( f \):
\[ f' = \frac{25}{9} f \]

### Final Answer
The frequency of \( K_\alpha \) X-ray emitted from the element with atomic number 51 is:
\[ f' = \frac{25}{9} f \]