Verified Solution • 2025–26 Session moderate

QPages 393-394: Q4: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 393-394: Q4 Solution: If the potential difference applied to the tube is doubled and…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume that increasing the filament‑target distance changes the X‑ray cutoff because they confuse electric field with potential difference; however, the minimum wavelength $lambda_{\min}=\frac{hc}{eV}$ depends only on the accelerating voltage $V$, not on the electrode spacing. Doubling the distance while also doubling $V$ halves $lambda_{\min}$, regardless of the gap.

If the potential difference applied to the tube is doubled and the separation between the filament and the target is also doubled, the cutoff wavelength

Multiple Choice Options

  • A will remain unchanged
  • B will be doubled
  • C will be halved
  • D will become four times the original

Step-by-Step Verified Solution

**Answer:** (c) will be halved. **Step-by-Step Explanation:** __min = hc/eV. If V is doubled, __min is halved. The separation doesn't matter.

All Questions in this Chapter

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