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JEE Main 2026 (21 Jan, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2026 (21 Jan, Shift 2)

Q10JEE Main 2026 · 21 JAN · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

A particle having electric charge 3×10−19 C3 \times 10^{-19}\text{ C} and mass 6×10−27 kg6 \times 10^{-27}\text{ 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\alpha \times 10^{-12}\text{ m}. The value of α\alpha is ______. (Take Planck's constant =6.6×10−34 J s= 6.6 \times 10^{-34}\text{ J s})

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. λ=h2mqV\lambda = \dfrac{h}{\sqrt{2mqV}}
  2. 2mqV=2×6×10−27×3×10−19×1.21=43.56×10−462mqV = 2 \times 6 \times 10^{-27} \times 3 \times 10^{-19} \times 1.21 = 43.56 \times 10^{-46}, so 2mqV=6.6×10−23\sqrt{2mqV} = 6.6 \times 10^{-23}
  3. λ=6.6×10−346.6×10−23=10−11 m=10×10−12 m\lambda = \dfrac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}} = 10^{-11}\text{ m} = 10 \times 10^{-12}\text{ m}
  4. So α=10\alpha = 10.
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