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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Dual Nature of Matter

JEE Main 2019 · 8 April · Morning Shift · Physics · Dual Nature of Matter

Two particles move at right angle to each other. Their de Broglie...

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:

Options

  1. A2λ=1λ1+1λ2\dfrac{2}{\lambda} = \dfrac{1}{{\lambda}_{1}} + \dfrac{1}{{\lambda}_{2}}
  2. Bλ=λ1+λ22\lambda = \dfrac{{\lambda}_{1} + {\lambda}_{2}}{2}
  3. Cλ=λ1λ2\lambda = \sqrt{{\lambda}_{1}{\lambda}_{2}}
  4. D1λ2=1λ12+1λ22\dfrac{1}{{\lambda}^{2}} = \dfrac{1}{{\lambda}_{1}^{2}} + \dfrac{1}{{\lambda}_{2}^{2}}

Correct answer

D. 1λ2=1λ12+1λ22\dfrac{1}{{\lambda}^{2}} = \dfrac{1}{{\lambda}_{1}^{2}} + \dfrac{1}{{\lambda}_{2}^{2}}

Step-by-step solution

\overset{\to}{{P}_{1}} = \dfrac{h}{{\lambda}_{1}}\overset{^}{i}\text{,}\overset{\to}{{P}_{2}} = \dfrac{h}{{\lambda}_{2}}\overset{^}{j}<br/><br/>\overset{\to}{P} = \overset{\to}{{P}_{1}} + \overset{\to}{{P}_{2}} = \dfrac{h}{{\lambda}_{1}}\overset{^}{i} + \dfrac{h}{{\lambda}_{2}}\overset{^}{j}<br/>∣P→∣=h1λ12+1λ22<br/>\left|\overset{\to}{P}\right| = h\sqrt{\dfrac{1}{{\lambda}_{1}^{2}} + \dfrac{1}{{\lambda}_{2}^{2}}}
hλ=h1λ12+1λ22\dfrac{h}{\lambda} = h\sqrt{\dfrac{1}{{\lambda}_{1}^{2}} + \dfrac{1}{{\lambda}_{2}^{2}}}
1λ2=1λ12+1λ22\dfrac{1}{{\lambda}^{2}} = \dfrac{1}{{\lambda}_{1}^{2}} + \dfrac{1}{{\lambda}_{2}^{2}}
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