Physics · Atomic Physics

JEE Advanced 2024 — Paper 2 — Question 15

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

Option A:
2.53
Correct
Option B:
1.27
Option C:
2.24
Option D:
1.58

Answer: A

Step-by-step solution

$\quad \frac{(\mathrm{K} \alpha)_{\mathrm{Z}=46}}{\lambda_{\text {cutoff }}}=2$

\frac{(\mathrm{K} \alpha)_{\mathrm{Z}=41}}{\lambda_{\text {cutoff }}}=\mathrm{x} \end{gathered}$$ Dividing both equation $\frac{1 /(46-1)^{2}}{1 /(41-1)^{2}}=\frac{2}{x}$ $x=2.53$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Physics
Chapter: Atomic Physics
Topic: X-Rays