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Two particles move at right angle to each other. Their de Broglie wavelengths are $λ_{1} a n d λ_{2}$
respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength $λ$ of the final particle is given by

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$\frac{1}{λ^{2}} = \frac{1}{λ_{1}^{2}} + \frac{1}{λ_{2}^{2}}$

$λ = \sqrt{λ_{1}^{} λ_{2}}$

$λ = \frac{λ_{1}^{} + λ_{2}}{2}$

$\frac{2}{λ} = \frac{1}{λ_{1}^{}} + \frac{1}{λ_{2}}$

answer is A.

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Momentum of single particle $\bar{P} = P_{1} \hat{i} + P_{2} \hat{j} (s a y)$

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