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QPages 395-396: Q5: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q5 Solution: The X-ray coming from a Coolidge tube has a cutoff wavelength…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug λ = 80 pm directly into $K=hc/\lambda$ using $h$ in J·s or $hc$ in eV·s without converting pm to the same length unit, producing a factor‑1000 error. Always express λ in metres when using $h=6.63\times10^{-34}\,$J·s, or use $hc=1240\,$eV·nm and convert pm to nm first.

The X-ray coming from a Coolidge tube has a cutoff wavelength of 80 pm. Find the kinetic energy of the electrons hitting the target. 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:** Kinetic energy = 15.5 keV. **Step-by-Step Explanation:** Given: __min = 80 pm = 80 _ 10___ m. Kinetic energy of electrons: K.E. = hc/__min = (4.14 _ 10___ _ 3 _ 10_) / (80 _ 10___) eV = 1.55 _ 10_ eV = 15.5 keV [citation:8].

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