The de Broglie wavelength is λ=mvh where v is found from energy conservation.
From qV=21mv2: v=m2qV=6×10−272×3×10−19×1.21=1.21×108=1.1×104 m/s.
Therefore: λ=6×10−27×1.1×1046.6×10−34=6.6×10−236.6×10−34=10−11 m = 10×10−12 m.
Thus α=10.
A particle having electric charge 3 × 10^-19 C and mass 6 × 10^-27 ~kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the…
JEE Main 2026 — Physics Modern Physics
2026integermedium
A particle having electric charge 3×10−19C and mass 6×10−27 kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α×10−12 m. The value of α is ____ - (Take Planck's constant =6.6×10−34 J.s)
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Held on 21 Jan 2026 · Verified 6 Jul 2026.
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