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QPages 395-396: Q4: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q4 Solution: What potential difference should be applied across an X-ray…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often read “wavelength not less than 0.10 nm” as “λ ≤ 0.10 nm” and then use a larger voltage than required; the correct interpretation is λ ≥ 0.10 nm, so the minimum accelerating voltage is V = hc/(e λ) with λ = 0.10 nm. Always translate the verbal inequality into the proper mathematical bound before applying the photon‑energy formula.

What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule? 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:** Potential difference = 12.4 kV, Maximum energy = 2 _ 10___ J. **Step-by-Step Explanation:** Given: __min = 0.10 nm. Potential difference: V = hc/e__min = (4.14 _ 10___ _ 3 _ 10_) / (0.10 _ 10__) = 12.4 _ 10_ V = 12.4 kV [citation:8]. Maximum energy: E = eV = 1.6 _ 10___ _ 12.4 _ 10_ = 1.98 _ 10___ J � 2 _ 10___ J [citation:8].

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