JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

A particle having electric charge $3 \times 1 0^{- 19}$ C and mass $6 \times 1 0^{- 27}$ kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is $α \times 1 0^{- 12}$ m. The value of $α$ is _________.

(Take Planck's constant = $6.6 \times 1 0^{- 34}$ J.s)

Your answer

Solution

For a charge accelerated through potential V, kinetic energy $= q V$ , and momentum $p = 2 m q V$ . de Broglie wavelength: $λ = \frac{h}{2 m q V}$ .

Compute the denominator: $2 m q V = 2 \times (6 \times 1 0^{- 27}) \times (3 \times 1 0^{- 19}) \times 1.21 = 4.356 \times 1 0^{- 45}$ . So $2 m q V \approx 6.6 \times 1 0^{- 23}$ .

$λ = \frac{6.6 \times 1 0 ^{- 34}}{6.6 \times 1 0 ^{- 23}} = 1 0^{- 11}$ m $= 10 \times 1 0^{- 12}$ m.

Therefore $α = 10$ .

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