Verified Solution • 2025–26 Session moderate

QPages 395-396: Q9: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q9 Solution: The electron beam in a colour TV is accelerated through 32 kV…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the accelerating voltage (32 kV) directly into $\lambda=hc/eV$ without converting it to electron‑volts, or they mix $h$ in joule‑seconds with $V$ in volts, missing the factor $e=1.6\times10^{-19}\,$C. Always express the photon energy as $E=eV$ (or use $h$ in eV·s) before applying $\lambda=hc/E$ to avoid a 10³ error.

The electron beam in a colour TV is accelerated through 32 kV and then strikes the screen. What is the wavelength of the most energetic X-ray photon? 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:** Wavelength = 38.8 pm. **Step-by-Step Explanation:** Given: V = 32 kV. __min = hc/eV = (4.14 _ 10___ _ 3 _ 10_) / (32 _ 10_) = 3.88 _ 10___ m = 38.8 pm [citation:1].

All Questions in this Chapter

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