A metal target with atomic number $Z=46$ is bombarded with a high energy electron beam. The emission of…

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. 2.53
  2. 1.27
  3. 2.24
  4. 1.58

Solution

\(\begin{aligned} & \frac{1}{\lambda_\alpha}=\frac{3}{4} R(Z-1)^2 p \\ & \lambda_{\text {cut }}=\frac{h c}{e V} \\ & \Rightarrow \text { Ratio } \propto \frac{1}{(Z-1)^2} \text { for same beam } \\ & \frac{Z}{x}=\frac{40^2}{45^2} \\ & \Rightarrow x=\frac{45^2}{40^2} .2 \approx 2.53 \end{aligned}\)

Asked in: JEE Advanced 2024 (Paper 2)

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