Two elements A and B with atomic numbers `Z_(A)` and `Z_(B)` are used to produce charateristic X-rays with frequencies `v_(A)` and `v_(B)` respectively. If `Z_(A):Z_(B)=1:2,` then `v_(A):v_(B)` will be

Options

A. `1:sqrt(2)`
B. `1:8`
C. `4:1`
D. `1:4`

Explanation

To solve the problem, we need to find the ratio of the frequencies of characteristic X-rays produced by two elements A and B, given their atomic numbers and the relationship defined by Moseley's law.

### Step-by-Step Solution:

1. **Understanding Moseley's Law**:
Moseley's law states that the frequency (ν) of the characteristic X-rays emitted by an element is proportional to the square of its atomic number (Z). Mathematically, this can be expressed as:
\[ \nu \propto Z^2 \]
This means that the frequency of the X-rays is directly related to the square of the atomic number of the element.

2. **Setting Up the Ratios**:
Given that the atomic numbers of elements A and B are \( Z_A \) and \( Z_B \) respectively, we can express the frequencies of the X-rays as:
\[ \nu_A = K \cdot Z_A^2 \]
\[ \nu_B = K \cdot Z_B^2 \]
where \( K \) is a constant.

3. **Finding the Ratio of Frequencies**:
We need to find the ratio \( \frac{\nu_A}{\nu_B} \):
\[ \frac{\nu_A}{\nu_B} = \frac{K \cdot Z_A^2}{K \cdot Z_B^2} = \frac{Z_A^2}{Z_B^2} \]

4. **Using the Given Ratio of Atomic Numbers**:
We are given that the ratio of the atomic numbers is:
\[ \frac{Z_A}{Z_B} = \frac{1}{2} \]
Squaring both sides gives:
\[ \left(\frac{Z_A}{Z_B}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

5. **Final Ratio of Frequencies**:
Therefore, substituting this back into our frequency ratio:
\[ \frac{\nu_A}{\nu_B} = \frac{1}{4} \]
This means that the ratio of the frequencies \( \nu_A : \nu_B \) is:
\[ \nu_A : \nu_B = 1 : 4 \]

### Conclusion:
The final answer is:
\[ \nu_A : \nu_B = 1 : 4 \]