QJEEAdv24Ph29ab: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph29ab Solution: A metal target with atomic number is bombarded with a high…

To solve this problem, we need to understand the relationship between the atomic number $Z$ , the characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.

The cut-off wavelength, also known as the minimum wavelength ( $\lambda_{\text{min}}$ ), corresponds to the maximum energy of the X-rays produced and can be related to the accelerating voltage used in the X-ray tube. The characteristic X-rays, such as the $K_\alpha$ -line, correspond to specific electron transitions in the atom and are dependent on the atomic number $Z$ of the target material.

The wavelength of the $K_\alpha$ -line is given by Moseley's Law, which can be approximated as:

$\lambda_{K_\alpha} \propto \frac{1}{(Z - 1)^2}$

The cut-off wavelength is given by the relationship between the energy of the incident electrons and the X-ray produced:

$\lambda_{\text{min}} = \frac{hc}{eV}$

Now, given the ratio $r$ for the first metal target with $Z = 46$ :

$r = \frac{\lambda_{K_\alpha}}{\lambda_{\text{min}}} = 2$

For the second metal target with $Z = 41$ , let's denote the new ratio as $r_2$ . We assume the same cutoff wavelength since the same electron beam is used.

The ratio for $Z = 41$ can be calculated using the relationship between the $K_\alpha$ -line wavelenghts:

$\frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{Z_2 - 1}{Z_1 - 1} \right)^2$

Substituting $Z_1 = 46$ and $Z_2 = 41$ :

$\frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{41 - 1}{46 - 1} \right)^2 = \left( \frac{40}{45} \right)^2 = \left( \frac{8}{9} \right)^2 = \frac{64}{81}$

Let $\lambda_{K_\alpha (1)} = 2 \lambda_{\text{min}}$ because $r = 2$ . Therefore:

$\lambda_{K_\alpha (2)} = \lambda_{K_\alpha (1)} \cdot \frac{81}{64} = 2 \lambda_{\text{min}} \cdot \frac{81}{64}$

Thus, the new ratio $r_2$ is:

$r_2 = \frac{\lambda_{K_\alpha (2)}}{\lambda_{\text{min}}} = 2 \cdot \frac{81}{64} = \frac{162}{64} = 2.53$

So, the value of $r$ for the second metal target with $Z = 41$ is:

Option A: 2.53