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# Engineering

A particle 'P' is formed due to a completely inelastic collision of particles 'x'\:\: and \:\:'y' having de-broglie wavelength '\lambda_{x}'\:\:and\:\:'\lambda_{y}' respectively . If x\:\:and\:\:y were moving in opposite directions , then the de-broglie wavelength of 'P' is :

 

 

 

  • Option 1)

    \frac{\lambda _{x}\lambda _{y}}{\lambda _{x}+\lambda _{y}}

     

  • Option 2)

    \frac{\lambda_{x}\lambda_{y}}{\left | \lambda_{x}-\lambda_{y} \right |}

  • Option 3)

    \lambda _{x}-\lambda _{y}

  • Option 4)

    \lambda _{x}+\lambda _{y}

 

Answers (1)
S solutionqc

 

\\m_{x}V_{x}-m_{y}V_{y}=(m_{x}+m_{y})V_{c}\\\\\:\Rightarrow \frac{h}{\lambda_{x}}-\frac{h}{\lambda_{y}}=\frac{h}{\lambda^{'}}

\Rightarrow \lambda ^{'}=\frac{\lambda_{x}\lambda_{y}}{\left | \lambda_{x}-\lambda_{y} \right |}

 


Option 1)

\frac{\lambda _{x}\lambda _{y}}{\lambda _{x}+\lambda _{y}}

 

Option 2)

\frac{\lambda_{x}\lambda_{y}}{\left | \lambda_{x}-\lambda_{y} \right |}

Option 3)

\lambda _{x}-\lambda _{y}

Option 4)

\lambda _{x}+\lambda _{y}

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