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

 

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