Verified Solution • 2025–26 Session moderate

QPages 395-396: Q10: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q10 Solution: When 40 kV is applied across an X-ray tube, X-ray is obtained…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often set the photon energy equal to the applied voltage directly ($h\nu = V$) and omit the factor of the elementary charge. Remember that the kinetic energy of the electrons is $eV$ (in joules), so the correct relation is $h\nu_{\max}=eV$; convert the voltage to energy before dividing by the frequency.

When 40 kV is applied across an X-ray tube, X-ray is obtained with a maximum frequency of 9.7×1018 Hz9.7 \times 10^{18} \text{ Hz}. Calculate the value of Planck constant from these data. Use Planck constant h=6.63×10−34 Js=4.14×10−15 eVsh = 6.63 \times 10^{-34} \text{ Js} = 4.14 \times 10^{-15} \text{ eVs}, speed of light c=3×108 m/sc = 3 \times 10^8 \text{ m/s}.)

Step-by-Step Verified Solution

**Answer:** Planck's constant h = 4.14 _ 10___ eV-s. **Step-by-Step Explanation:** Given: V = 40 kV, __max = 9.7 _ 10__ Hz. Energy: eV = h__max. h = eV/__max = (1.6 _ 10___ _ 40 _ 10_) / (9.7 _ 10__) = 6.6 _ 10___ J-s = 4.14 _ 10___ eV-s [citation:1].

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