JEE Advanced · Physics · 23. EM Waves
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
- A 2.53
- B 1.27
- C 2.24
- D 1.58
Answer & Solution
Correct Answer
(A) 2.53
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{1}{\lambda_\alpha}=\frac{3}{4} R(Z-1)^2 p \\ & \lambda_{\text {cut }}=\frac{h c}{e V} \\ & \Rightarrow \text { Ratio } \propto \frac{1}{(Z-1)^2} \text { for same beam } \\ & \frac{Z}{x}=\frac{40^2}{45^2} \\ & \Rightarrow x=\frac{45^2}{40^2} .2 \approx 2.53 \end{aligned}\)
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