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QPage 363: Q12: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Page 363: Q12 Solution: Consider the de-Broglie wavelength of an electron and a proton. …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that equal kinetic energy implies equal de-Broglie wavelength, failing to recognize that $\lambda = h/\sqrt{2mE}$ depends on mass, whereas $\lambda = h/p$ is mass-independent when momentum is fixed.

Consider the de-Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have a) the same speed b) the same momentum c) the same energy?

Step-by-Step Verified Solution

**Answer:** (a) Proton has smaller wavelength. (b) Both have the same wavelength. (c) Proton has smaller wavelength. **Step-by-Step Explanation:** de Broglie wavelength λ=h/p=h/(mv)\lambda = h/p = h/(mv). (a) Same speed: λ∝1/m\lambda \propto 1/m. Proton is heavier, so its wavelength is smaller. (b) Same momentum: λ=h/p\lambda = h/p, so both have the same wavelength. (c) Same energy: E=p2/2mE = p^2/2m, so p=2mEp = \sqrt{2mE}. λ=h/2mE\lambda = h/\sqrt{2mE}. Proton is heavier, so its wavelength is smaller.

All Questions in this Chapter

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