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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 , speed of light .)
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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