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JEE Main 2019 (8 Apr, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2019 (8 Apr, Shift 1)

Q126JEE Main 2019 · 8 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength

Two particles move at right angle to each other. Their de-Broglie wavelengths are λ1\lambda_1 and λ2\lambda_2 respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength λ\lambda of the final particle, is given by:

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Answer: (b)

p₁ = h/λ₁p₂p = h/λ
  1. Momentum is conserved; p⃗1⊥p⃗2\vec p_1 \perp \vec p_2, so p2=p12+p22p^2 = p_1^2 + p_2^2.
  2. With p=h/λp = h/\lambda: 1λ2=1λ12+1λ22\dfrac{1}{\lambda^2} = \dfrac{1}{\lambda_1^2} + \dfrac{1}{\lambda_2^2}
  3. So the answer is (b).
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