Verified Solution • 2025–26 Session moderate

QPages 395-396: Q7: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q7 Solution: The distance between the cathode filament) and the target in an …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often drop the $10^{-12}$ factor when converting $30\,\text{pm}$ to metres, treating it as $30\times10^{-9}$ m or ignoring the exponent altogether; this leads to a voltage off by three orders of magnitude. Always write $\lambda=30\,\text{pm}=30\times10^{-12}\,\text{m}=3.0\times10^{-11}\,\text{m}$ and keep the powers of ten consistent when using $h$ in eV·s.

The distance between the cathode filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30 pm, find the electric field between the cathode and 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:** Electric field = 2.76 _ 10_ V/m. **Step-by-Step Explanation:** Given: d = 1.5 m, __min = 30 pm = 30 _ 10___ m. V = hc/e__min = (4.14 _ 10___ _ 3 _ 10_) / (30 _ 10___) = 4.14 _ 10_ V. Electric field: E = V/d = (4.14 _ 10_) / 1.5 = 2.76 _ 10_ V/m [citation:8].

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