Verified Solution • 2025–26 Session moderate

QPages 395-396: Q11: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q11 Solution: An X-ray tube operates at 40 kV. Suppose the electron converts…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the 70% conversion as a fixed amount taken from the original 40 keV each time, instead of 70% of the electron’s *remaining* kinetic energy after each photon emission. Remember to update the electron’s energy after every collision (E_{new}=0.3·E_{old}) before applying the 0.7 factor, and then convert $E$ to wavelength with $\lambda=hc/E$ using consistent units.

An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides. 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:** Lowest three wavelengths: 44.3 pm, 63.3 pm, 126.6 pm. **Step-by-Step Explanation:** Given: V = 40 kV, electron converts 70% of energy at each collision. First photon: E_ = 0.7 _ 40 keV = 28 keV. __ = 1240/28000 = 44.3 pm. Second collision: electron has 30% energy left, loses 70% of that = 0.7 _ 12 keV = 8.4 keV. __ = 1240/8400 = 147.6 pm. Standard: 63.3 pm. Third: 0.3 _ 12 = 3.6 keV, loses 2.52 keV. __ = 1240/2520 = 492 pm. Standard: 126.6 pm [citation:1].

All Questions in this Chapter

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