Iron (z = 26) emits `k_(alpha)` line of wavelength `1.96A^(@)`. For an element of unknown atomic number the wavelength of `k_(alpha)` line is `0.49A^(0)` The atomic number of the unknown element is

Options

A. 51
B. 49
C. 196
D. 31

Answer

To find the atomic number of the unknown element based on the given wavelengths of the K-alpha lines, we can use the relationship between the wavelength of the K-alpha line and the atomic number.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- For Iron (Z = 26), the wavelength of the K-alpha line (λ1) = 1.96 Å.
- For the unknown element, the wavelength of the K-alpha line (λ2) = 0.49 Å.

2. **Use the Relationship:**
The relationship between the wavelength (λ) and atomic number (Z) for K-alpha lines can be expressed as:
\[ \frac{1}{\lambda} \propto (Z - b)^2 \]
For K-alpha lines, \( b = 1 \). Therefore, we can write:
\[ \frac{1}{\lambda} \propto (Z - 1)^2 \]

3. **Set Up the Ratio:**
We can set up a ratio for the two elements:
\[ \frac{\lambda_2}{\lambda_1} = \frac{(Z_1 - 1)^2}{(Z_2 - 1)^2} \]
Substituting the known values:
\[ \frac{0.49}{1.96} = \frac{(26 - 1)^2}{(Z_2 - 1)^2} \]

4. **Simplify the Ratio:**
Calculate the left side:
\[ \frac{0.49}{1.96} = \frac{1}{4} \]
Therefore, we have:
\[ \frac{1}{4} = \frac{25^2}{(Z_2 - 1)^2} \]

5. **Cross-Multiply:**
Cross-multiplying gives:
\[ (Z_2 - 1)^2 = 25^2 \times 4 \]
\[ (Z_2 - 1)^2 = 625 \times 4 = 2500 \]

6. **Take the Square Root:**
Taking the square root of both sides:
\[ Z_2 - 1 = \sqrt{2500} = 50 \]

7. **Solve for Z2:**
Adding 1 to both sides:
\[ Z_2 = 50 + 1 = 51 \]

8. **Conclusion:**
The atomic number of the unknown element is **51**.