The wavelength of `K_(alpha)` line for an element of atomic number `43 is lambda`. Then the wavelength of `K_(alpha)` line for an element of atomic number `29` is

Options

A. `(43)/(29) lambda`
B. `(42)/(28) lambda`
C. `(9)/(4) lambda`
D. `(4)/(9) lambda`

Explanation

To solve the problem, we will use Moseley's law, which relates the wavelength of the K-alpha line to the atomic number of the element. The law states that the frequency of the emitted X-rays is proportional to the square of the atomic number (Z) of the element.

### Step-by-Step Solution:

1. **Understanding Moseley's Law**:
Moseley's law can be expressed as:
\[ \nu = k(Z - b)^2 \]
where \( \nu \) is the frequency of the emitted X-ray, \( Z \) is the atomic number, \( b \) is a constant (which is typically taken as 1 for K-alpha transitions), and \( k \) is a proportionality constant.

2. **Relating Frequency and Wavelength**:
The relationship between frequency (\( \nu \)) and wavelength (\( \lambda \)) is given by:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light. Therefore, we can express the wavelength in terms of the atomic number:
\[ \lambda = \frac{c}{k(Z - b)^2} \]

3. **Finding the Wavelength for Atomic Number 43**:
Let’s denote the wavelength of the K-alpha line for the element with atomic number 43 as \( \lambda_{43} \):
\[ \lambda_{43} = \frac{c}{k(43 - 1)^2} = \frac{c}{k(42)^2} \]
Given that \( \lambda_{43} = \lambda \), we have:
\[ \lambda = \frac{c}{k(42)^2} \]

4. **Finding the Wavelength for Atomic Number 29**:
Now, we will find the wavelength for the element with atomic number 29:
\[ \lambda_{29} = \frac{c}{k(29 - 1)^2} = \frac{c}{k(28)^2} \]

5. **Finding the Ratio of Wavelengths**:
To find the relationship between \( \lambda \) and \( \lambda_{29} \), we can set up the ratio:
\[ \frac{\lambda_{29}}{\lambda} = \frac{c/(k(28)^2)}{c/(k(42)^2)} = \frac{(42)^2}{(28)^2} \]
Simplifying this gives:
\[ \frac{\lambda_{29}}{\lambda} = \left(\frac{42}{28}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]
Therefore, we can express \( \lambda_{29} \) in terms of \( \lambda \):
\[ \lambda_{29} = \frac{9}{4} \lambda \]

### Final Answer:
The wavelength of the K-alpha line for the element with atomic number 29 is:
\[ \lambda_{29} = \frac{9}{4} \lambda \]