The wavelength of `K_(alpha)` X-rays of two metals `A and B` are` 4 // 1875 R and 1// 675 R`, respectively , where `R` is rydberg 's constant. The number of electron lying between `A and B` according to this lineis

Options

A. 3
B. 6
C. 5
D. 4

Explanation

To solve the problem, we need to determine the number of electrons (or atomic numbers) lying between two metals A and B based on their Kα X-ray wavelengths. The wavelengths are given as \( \lambda_A = \frac{4}{1875} R \) and \( \lambda_B = \frac{1}{675} R \), where \( R \) is the Rydberg constant.

### Step-by-Step Solution:

1. **Use the Rydberg Formula for Kα X-rays**:
The Rydberg formula for Kα X-rays is given by:
\[ \frac{1}{\lambda} = R \left( Z - 1 \right)^2 \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
For Kα transition, \( n_1 = 1 \) and \( n_2 = 2 \).

2. **Calculate for Metal A**:
Substitute the wavelength for metal A:
\[ \frac{1}{\lambda_A} = \frac{1}{\frac{4}{1875} R} = \frac{1875}{4R} \]
Plugging this into the Rydberg formula:
\[ \frac{1875}{4R} = R \left( Z_1 - 1 \right)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Simplifying the right-hand side:
\[ \frac{1875}{4R} = R \left( Z_1 - 1 \right)^2 \left( 1 - \frac{1}{4} \right) = R \left( Z_1 - 1 \right)^2 \left( \frac{3}{4} \right) \]
Cancel \( R \) from both sides:
\[ \frac{1875}{4} = \left( Z_1 - 1 \right)^2 \left( \frac{3}{4} \right) \]
Rearranging gives:
\[ \left( Z_1 - 1 \right)^2 = \frac{1875}{3} = 625 \]
Taking the square root:
\[ Z_1 - 1 = 25 \implies Z_1 = 26 \]

3. **Calculate for Metal B**:
Now substitute the wavelength for metal B:
\[ \frac{1}{\lambda_B} = \frac{1}{\frac{1}{675} R} = 675 R \]
Plugging this into the Rydberg formula:
\[ 675 R = R \left( Z_2 - 1 \right)^2 \left( \frac{3}{4} \right) \]
Cancel \( R \):
\[ 675 = \left( Z_2 - 1 \right)^2 \left( \frac{3}{4} \right) \]
Rearranging gives:
\[ \left( Z_2 - 1 \right)^2 = 675 \times \frac{4}{3} = 900 \]
Taking the square root:
\[ Z_2 - 1 = 30 \implies Z_2 = 31 \]

4. **Determine the Number of Electrons Between A and B**:
Now, we have:
- \( Z_1 = 26 \)
- \( Z_2 = 31 \)

The atomic numbers between 26 and 31 are 27, 28, 29, and 30. Thus, the number of electrons (or atomic numbers) between A and B is:
\[ 31 - 26 - 1 = 4 \]

### Final Answer:
The number of electrons lying between metals A and B is **4**.