JEE Advanced 2024 · Paper 2 · Physics Q3

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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 $r$ 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

2.53

1.27

2.24

1.58

Answer and worked solution

Official answer: A

The cut-off wavelength $\lambda_{min}=\frac{hc}{eV}$ depends only on the accelerating voltage, so it is the same for both targets.

By Moseley's law, $\lambda_{K\alpha}\propto\frac{1}{(Z-1)^2}$ . Therefore
$r'=r\cdot\frac{(46-1)^2}{(41-1)^2}=2\cdot\frac{2025}{1600}=2.53.$

Topics:Moseley's law X-rays

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