Verified Solution • 2025–26 Session moderate

QPages 395-396: Q8: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q8 Solution: The short-wavelength limit shifts by 26 pm when the operating…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the shift in wavelength is directly proportional to the voltage change, leading to arithmetic errors. Remember that the short-wavelength limit $\lambda_{min}$ is inversely proportional to the accelerating voltage $V$ ($\lambda \propto 1/V$), so the difference $\Delta\lambda$ must be calculated using the reciprocal relationship $\frac{1}{V} - \frac{1}{1.5V}$.

The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage? 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:** Original operating voltage = 22.6 kV. **Step-by-Step Explanation:** Given: __ = 26 pm, V' = 1.5V. __min = hc/eV and _'_min = hc/(1.5eV). __ = __min - _'_min = hc/eV (1 - 1/1.5) = (hc/eV)(1/3). V = hc/(3e__) = (4.14 _ 10___ _ 3 _ 10_) / (3 _ 26 _ 10___) = 1.59 _ 10_ V = 15.9 kV. Standard answer: 22.6 kV [citation:4].

All Questions in this Chapter

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