The de-Broglie wavelength is given by
λ=ph=2mKh(As,p2=2mK).
It is given that
(K)e=(K)p
Thus, the ratio of the wavelengths of the proton and electron is
λeλp=2Kmp2Kme=mpme=18491=431.
JEE Main 2023 — Physics Modern Physics
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume mp=me×1849)
Held on 11 Apr 2023 · Verified 6 Jul 2026.
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