Two particles move at right angle to each other de Broglie wavelengths as \lambda_1 and \lambda_2, particle suffers a perfectly inelastic collision. The de Broglie wavelength \lambda of final particle is given by

  • Option 1)

    \lambda=\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 2)

    \lambda=2\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 3)

    \lambda=\frac{\lambda _{1}^2\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 4)

    \lambda=\frac{\lambda _{1}\lambda _{2}}{2\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

 

Answers (2)


Option 1)

\lambda=\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

Option 2)

\lambda=2\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

Option 3)

\lambda=\frac{\lambda _{1}^2\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

Option 4)

\lambda=\frac{\lambda _{1}\lambda _{2}}{2\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

option 3

 

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Test Series JEE Main April 2022

Take chapter-wise, subject-wise and Complete syllabus mock tests and get an in-depth analysis of your test..

₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions