Verified Solution • 2025–26 Session moderate

QPages 395-396: Q3: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q3 Solution: Find the cutoff wavelength for the continuous X-rays coming…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug $h=6.63\times10^{-34}\,\text{J·s}$ and $V=30\,\text{kV}$ directly into $\lambda_{\min}=\frac{hc}{eV}$, forgetting that $eV$ is an energy unit; either use $h$ in eV·s or convert $V$ to joules by multiplying with the electron charge $e$. Mixing J and eV yields a wavelength off by a factor of $1.6\times10^{-19}$.

Find the cutoff wavelength for the continuous X-rays coming from an X-ray tube operating at 30 kV. Use Planck constant h=4.14×10−15 eVsh = 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:** Cutoff wavelength = 41.4 pm. **Step-by-Step Explanation:** Given: V = 30 kV = 30 _ 10_ V. Cutoff wavelength: __min = hc/eV = (4.14 _ 10___ eV-s _ 3 _ 10_ m/s) / (30 _ 10_ V) = 4.14 _ 10___ m = 41.4 pm [citation:8][citation:16].

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