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JEE · NEET Previous Year Question

$The de Broglie wavelength of an electron accelerated between two plates having a potential difference of 900 V is nearly$

Topic: JEE PYQ

Answer: D. Full step-by-step solution below.

A $0.015 nm$
B $0.01 nm$
C $0.02 nm$
D $0.04 nm$

Correct answer: D

Solution

(d) de Broglie wavelength is given by λ = \frac{12.27 \times 10 ^{- 10}}{V} = \frac{12.27 \times 10 ^{- 10}}{900} = \frac{12.27 \times 10 ^{- 10}}{30} = 0.04 nm

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