JEE Advanced 2024 Paper 2 · Q03 · X-rays
A metal target with atomic number $Z = 46$ is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio of the wavelengths of the $K_\alpha$-line and the cut-off is found to be $r = 2$. If the same electron beam bombards another metal target with $Z = 41$, the value of $r$ will be
Reveal answer + step-by-step solution
Correct answer:A
Solution
The cut-off (minimum) wavelength of the continuous X-ray spectrum is set by the accelerating potential and is independent of the target: $\lambda_{\text{cutoff}} = hc/(eV)$. For the same electron beam, $\lambda_{\text{cutoff}}$ is the same for both targets.
The $K_\alpha$-line wavelength follows Moseley's law: $\dfrac{1}{\lambda_{K_\alpha}} = R(Z-1)^2\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right) = \dfrac{3R}{4}(Z-1)^2$, so $\lambda_{K_\alpha} \propto \dfrac{1}{(Z-1)^2}$.
Hence $r = \dfrac{\lambda_{K_\alpha}}{\lambda_{\text{cutoff}}} \propto \dfrac{1}{(Z-1)^2}$.
Given $r_{Z=46} = 2$, so $r_{Z=41} = 2 \times \dfrac{(46-1)^2}{(41-1)^2} = 2 \times \dfrac{45^2}{40^2} = 2 \times \dfrac{2025}{1600} = 2 \times 1.265625 = 2.531$.
Units: $\lambda$ in metres on both sides; $r$ is dimensionless. Answer (A) $2.53$.
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