Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$\Delta \lambda$ 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
PhysicsElectromagnetic WavesJEE Main
Options:
  • A $\Delta \lambda^{\prime}=5 \Delta \lambda$
  • B $\Delta \lambda^{\prime}=4 \Delta \lambda$
  • C $\Delta \lambda^{\prime}=3 \Delta \lambda$
  • D $\Delta \lambda^{\prime} < 3 \Delta \lambda$
Solution:
2629 Upvotes Verified Answer
The correct answer is: $\Delta \lambda^{\prime}=3 \Delta \lambda$
$\begin{aligned} & \text { We know } \mathrm{eV}=\frac{h c}{\lambda} \Rightarrow \frac{V}{V^{\prime}}=\frac{\Delta \lambda^{\prime}}{\Delta \lambda} \\ & \Rightarrow \quad \frac{V}{V / 3}=\frac{\Delta \lambda^{\prime}}{\Delta \lambda} \\ & \Rightarrow \quad \Delta \lambda^{\prime}=3(\Delta \lambda)\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.