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QPages 395-396: Q21: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q21 Solution: Find the maximum potential difference which may be applied…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often compare the electron energy only with the K‑shell binding (69.5 keV) and answer 69.5 kV, forgetting that an L‑shell vacancy (11.3 keV) will already produce characteristic L X‑rays. To avoid any K or L lines, the accelerating voltage must stay below the lowest relevant binding energy—in this case the L‑shell energy of 11.3 keV.

Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristic K or L X-ray. The energy levels of the tungsten atom with an electron knocked out are as follows. Cell containing vacancy | K | L | M; Energy in keV | 69.5 | 11.3 | 2.3.

Step-by-Step Verified Solution

**Answer:** Maximum potential difference = 11.3 kV. **Step-by-Step Explanation:** For no characteristic X-rays to be emitted, the electron energy must be less than the energy needed to knock out a K-shell electron (the most tightly bound). The highest energy required is to knock out a K-shell electron: EK=69.5E_K = 69.5 keV. Wait�the table gives EK=69.5E_K = 69.5 keV, EL=11.3E_L = 11.3 keV, EM=2.3E_M = 2.3 keV. The minimum energy to knock out an L-shell electron is 11.3 keV. If the electron energy is less than 11.3 keV, no L or K X-rays can be produced. Hence, the maximum potential difference is 11.3 kV.

All Questions in this Chapter

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