An electron and a proton both start from rest and are accelerated separately. The electron is accelerated through 1836 V. The proton is to be accelerated through such a potential that its de Broglie wavelength is exactly equal to that of the electron. Take mₚ = 1836mₑ. If the proton accelerating potential is N V, find N.
Physics · Photoelectric effect and dual nature of matter · debroglie duel nature of light
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Question

An electron and a proton both start from rest and are accelerated separately. The electron is accelerated through 1836 V. The proton is to be accelerated through such a potential that its de Broglie wavelength is exactly equal to that of the electron. Take mₚ = 1836mₑ. If the proton accelerating potential is N V, find N.

Physics · Photoelectric effect and dual nature of matter · debroglie duel nature of light

G1 Physics · Accepted Answer

Equal de Broglie wavelengths imply equal momentum magnitudes: pₑ = pₚ. For a particle accelerated from rest through potential V, p = √(2mqV). Both electron and proton have charge magnitude e. Therefore equal momentum requires mₑVₑ = mₚVₚ. Substitute Vₑ = 1836 V and mₚ = 1836mₑ. Thus, mₑ×1836 = 1836mₑ×Vₚ. Hence, Vₚ = 1 V. Therefore N = 1.