JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Modern Physics

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

)

Answer: 10

Solution

The kinetic energy gained is

K = q V

, so the momentum is

p = 2 m K = 2 m q V

and the de Broglie wavelength is

λ = \frac{h}{2 m q V}

. Here

2 m q V = 2 (6 \times 1 0^{- 27}) (3 \times 1 0^{- 19}) (1.21) = 4.356 \times 1 0^{- 45}

, whose square root is

6.6 \times 1 0^{- 23} kg m/s

. Thus

λ = \frac{6.6 \times 1 0 ^{- 34}}{6.6 \times 1 0 ^{- 23}} = 1.0 \times 1 0^{- 11} m = 10 \times 1 0^{- 12} m

, so

α = 10

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Integer answer

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