For a charged particle accelerated through potential V, the de Broglie wavelength is λ=2mVeh.
The wavelength is inversely proportional to m.
Therefore: λ2λ1=m1m2=14=2.
This means λ1=2λ2, so x = 2 (nearest integer).
JEE Main 2026 — Chemistry Physical Chemistry
Two positively charged particles m1 and m2 have been accelerated across the same potential difference of 200 keV as shown below.

[Given mass of m1=1 amu and m2=4 amu]
The deBroglie wavelength of m1 will be x times of m2. The value of x is ____ (nearest integer)
Held on 28 Jan 2026 · Verified 6 Jul 2026.
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