Verified Solution • 2025–26 Session moderate

QPages 395-396: Q12: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q12 Solution: The wavelength of X-ray of tungsten is 21.3 pm. It takes 11.3…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often set the minimum tube voltage equal to the $K_\alpha$ photon energy alone, overlooking that the incident electron must first eject a K‑shell electron. The required voltage must supply the K‑shell binding energy, which equals the L‑shell binding energy plus the $K_\alpha$ photon energy, so $V_{min}=E_{L}+E_{K\alpha}$.

The wavelength of KαK_\alpha X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of KαK_\alpha X-ray? 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:** Minimum accelerating voltage = 69.5 kV. **Step-by-Step Explanation:** Given: _(K_alpha) = 21.3 pm, E_L = 11.3 keV. Energy of K_alpha: E = hc/_ = 1240/(21.3 _ 10__) = 58.2 keV. E_K = E_L + E = 11.3 + 58.2 = 69.5 keV. Minimum voltage = 69.5 kV [citation:1].

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