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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

  1. A 2.53
  2. B 1.27
  3. C 2.24
  4. D 1.58
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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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