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QPages 393-394: Q10: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 393-394: Q10 Solution: The figure shows the intensity-wavelength relations of X-rays…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often equate higher intensity or broader curves with a larger accelerating voltage, but the decisive indicator is the shortest wavelength: λ_min=hc/eV, so the tube whose spectrum extends farther into the short‑wavelength side has the higher V. Likewise, the positions (not heights) of characteristic peaks reveal the target Z—higher Z shifts peaks to shorter wavelengths.

The figure shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for the tube A in which the potential difference between the target and the filament is VAV_A and the atomic number of the target material is ZAZ_A. These quantities are VBV_B and ZBZ_B for the other tube. Then,

Multiple Choice Options

  • A VA>VB,ZA>ZBV_A > V_B, Z_A > Z_B
  • B VA>VB,ZA<ZBV_A > V_B, Z_A < Z_B
  • C VA<VB,ZA>ZBV_A < V_B, Z_A > Z_B
  • D VA<VB,ZA<ZBV_A < V_B, Z_A < Z_B

Step-by-Step Verified Solution

**Answer:** (d) V_A < V_B, Z_A < Z_B. **Step-by-Step Explanation:** From the figure, tube B has a smaller cutoff wavelength, so V_B > V_A. Tube B also has characteristic peaks at higher frequencies, so Z_B > Z_A.

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