Verified Solution • 2025–26 Session moderate

QPages 363-364: Q14: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 363-364: Q14 Solution: A proton and an electron are accelerated by the same potential…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that de Broglie wavelength is inversely proportional to mass ($\lambda \propto 1/m$) or linearly related to kinetic energy, leading to wrong comparisons. Remember that for particles accelerated by the same potential $V$, the wavelength scales as $\lambda \propto 1/\sqrt{m}$, so the lighter electron has a significantly larger wavelength than the proton.

A proton and an electron are accelerated by the same potential difference. Let

\lambda_e

and

\lambda_p

denote the de Broglie wavelengths of the electron and the proton, respectively.

Multiple Choice Options

A $\lambda_e = \lambda_p$
B $\lambda_e < \lambda_p$
C $\lambda_e > \lambda_p$
D The relation between $\lambda_e$ and $\lambda_p$ depends on the accelerating potential difference.

Step-by-Step Verified Solution

**Answer:** (c) __e > __p. **Step-by-Step Explanation:** _ = h/�(2mqV). Electron has smaller mass, so larger wavelength.

All Questions in this Chapter

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