AP EAMCET · PHYSICS · Electromagnetic Waves
\(\Delta \lambda\) is the difference between the wavelength of \(k_\alpha\) line and the minimum wavelength of the continuous X-ray spectrum when the X-ray tube is operated at a voltage \(V\). If the operating voltage is changed to \(V / 3\), then the above difference is \(\Delta \lambda^{\prime}\). Then
- A \(\Delta \lambda^{\prime}=5 \Delta \lambda\)
- B \(\Delta \lambda^{\prime}=4 \Delta \lambda\)
- C \(\Delta \lambda^{\prime}=3 \Delta \lambda\)
- D \(\Delta \lambda^{\prime} < 3 \Delta \lambda\)
Answer & Solution
Correct Answer
(C) \(\Delta \lambda^{\prime}=3 \Delta \lambda\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { We know } \mathrm{eV}=\frac{h c}{\lambda} \Rightarrow \frac{V}{V^{\prime}}=\frac{\Delta \lambda^{\prime}}{\Delta \lambda} \\ & \Rightarrow \quad \frac{V}{V / 3}=\frac{\Delta \lambda^{\prime}}{\Delta \lambda} \\ & \Rightarrow \quad \Delta…
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