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# Engineering

Two particles move at right angle to each other. Their de Broglie wavelengths are \lambda _{1} and \lambda _{2} respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength \lambda, of the final particle, is given by :
 

  • Option 1)

    \lambda=\frac{\lambda_{1}+\lambda_{2}}{2}

  • Option 2)

    \lambda=\sqrt{\lambda_{1}+\lambda_{2}}

  • Option 3)

    \frac{1}{\lambda_{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}}

     

  • Option 4)

    \frac{2}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}

 

Answers (1)
S solutionqc

P_{1}\; \widehat{l}-P_{2}\; \widehat{J}=\vec{P}........\; \; \; \; \; (1)

We know P=\frac{h}{\lambda }

\Rightarrow \frac{h}{\lambda _{1}}\; \widehat{l}-\frac{h}{\lambda _{2}}\; \widehat{J}=\frac{h}{\lambda }

\Rightarrow \frac{h}{\lambda }=\left ( \sqrt{\frac{1}{\lambda {_{1}}^{2}}+\frac{1}{\lambda {_{2}}^{2}}} \right )*{h}

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

or \frac{1}{\lambda_{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}}


Option 1)

\lambda=\frac{\lambda_{1}+\lambda_{2}}{2}

Option 2)

\lambda=\sqrt{\lambda_{1}+\lambda_{2}}

Option 3)

\frac{1}{\lambda_{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}}

 

Option 4)

\frac{2}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}

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