# 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) $\frac{2}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$

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

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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