# 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) $\lambda _{x}+\lambda _{y}$

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

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

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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