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Question

The de-broglie wavelength of an electron accelerated by an electric potential of 'V' volts is given by
(Given:me=9.1×1031kg,e=1.6×1019C,h=6.6×1034J/s)

A
λ=1.23mm
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B
λ=1.23hm
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C
λ=1.23Vnm
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D
λ=1.23Vnm
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Solution

The correct option is C λ=1.23Vnm
From de-Broglie equation,
λ=hmv=h2mEk=h2meV
λ = wavelength
Ek = kinetic energy
m = mass of the electron
h = planck's constant
Subtitute values in above equation,
λ=6.6 × 10342 × 9.1 × 1031 × 1.6 × 1019 × Vλ=1.23V nm

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