Two elements X and Y have atomic number 15 and a respectively. The ratio of wavelength of `K_alpha` X-rays emitted by both is 1:4. Find the value of a

Explanation

To solve the problem, we need to find the atomic number \( a \) of element Y, given that the atomic number of element X is 15 and the ratio of the wavelengths of K-alpha X-rays emitted by both elements is 1:4.

### Step-by-step Solution:

1. **Understanding the relationship**: According to Moseley's law, the frequency \( \nu \) of the X-rays is related to the atomic number \( Z \) by the formula:
\[ \nu \propto (Z - b)^2 \]
where \( b \) is the screening constant. For K-alpha lines, \( b = 1 \).

2. **Expressing frequency in terms of wavelength**: The frequency is inversely proportional to the wavelength \( \lambda \):
\[ \nu \propto \frac{1}{\lambda} \]
Therefore, we can write:
\[ \lambda \propto \frac{1}{(Z - 1)^2} \]

3. **Setting up the ratio**: Given the ratio of the wavelengths of X-rays emitted by elements X and Y is \( \frac{\lambda_X}{\lambda_Y} = \frac{1}{4} \), we can express this in terms of their atomic numbers:
\[ \frac{\lambda_X}{\lambda_Y} = \frac{(Z_Y - 1)^2}{(Z_X - 1)^2} \]
Substituting \( Z_X = 15 \) and \( Z_Y = a \):
\[ \frac{(a - 1)^2}{(15 - 1)^2} = \frac{1}{4} \]

4. **Calculating the squares**: Simplifying the equation:
\[ \frac{(a - 1)^2}{14^2} = \frac{1}{4} \]
This implies:
\[ (a - 1)^2 = \frac{14^2}{4} \]
\[ (a - 1)^2 = \frac{196}{4} = 49 \]

5. **Taking the square root**: Taking the square root of both sides gives:
\[ a - 1 = \pm 7 \]
This results in two possible equations:
- \( a - 1 = 7 \) → \( a = 8 \)
- \( a - 1 = -7 \) → \( a = -6 \) (not a valid atomic number)

6. **Final answer**: Therefore, the only valid solution is:
\[ a = 8 \]

### Conclusion:
The value of \( a \) is \( 8 \).