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

Answer

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 K-alpha X-ray wavelengths. The wavelengths are given as:

- For metal A: \( \lambda_A = \frac{4}{1875} R \)
- For metal B: \( \lambda_B = \frac{1}{675} R \)

Where \( R \) is Rydberg's constant.

### Step 1: Use the Rydberg formula for K-alpha X-rays

The wavelength of the K-alpha X-rays can be related to the atomic number \( Z \) of the metal using the Rydberg formula:

\[ \frac{1}{\lambda} = R \left( Z - 1 \right)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For K-alpha transitions, \( n_1 = 1 \) and \( n_2 = 2 \). Thus, the formula simplifies to:

\[ \frac{1}{\lambda} = R (Z - 1)^2 \left( 1 - \frac{1}{4} \right) = R (Z - 1)^2 \left( \frac{3}{4} \right) \]

### Step 2: Calculate \( Z \) for metal A

For metal A, substituting \( \lambda_A \):

\[ \frac{1}{\lambda_A} = \frac{1875}{4} R \]

Equating this to the Rydberg formula:

\[ \frac{1875}{4} R = R (Z_A - 1)^2 \left( \frac{3}{4} \right) \]

Cancelling \( R \) from both sides:

\[ \frac{1875}{4} = (Z_A - 1)^2 \left( \frac{3}{4} \right) \]

Multiplying both sides by \( \frac{4}{3} \):

\[ 625 = (Z_A - 1)^2 \]

Taking the square root:

\[ Z_A - 1 = 25 \implies Z_A = 26 \]

### Step 3: Calculate \( Z \) for metal B

For metal B, substituting \( \lambda_B \):

\[ \frac{1}{\lambda_B} = 675 R \]

Equating this to the Rydberg formula:

\[ 675 = R (Z_B - 1)^2 \left( \frac{3}{4} \right) \]

Cancelling \( R \):

\[ 675 = (Z_B - 1)^2 \left( \frac{3}{4} \right) \]

Multiplying both sides by \( \frac{4}{3} \):

\[ 900 = (Z_B - 1)^2 \]

Taking the square root:

\[ Z_B - 1 = 30 \implies Z_B = 31 \]

### Step 4: Determine the number of electrons between metals A and B

Now we have:

- \( Z_A = 26 \)
- \( Z_B = 31 \)

The number of electrons (or atomic numbers) lying between A and B is:

\[ Z_A + 1, Z_A + 2, Z_A + 3, Z_A + 4 \quad \text{(which are 27, 28, 29, 30)} \]

Thus, the number of electrons between metals A and B is:

\[ 31 - 26 - 1 = 4 \]

### Final Answer

The number of electrons lying between metals A and B is **4**.

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