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

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

$λ = \sqrt{λ_{1} λ_{2}}$

$λ = \frac{λ_{2} + λ_{2}}{2}$

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

answer is A.

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Detailed Solution

From the de Broglie relation,

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$p_{1} = \frac{h}{λ_{1}}$

$p_{2} = \frac{h}{λ_{2}}$

Momentum of the final particle (p_f) is given by

$∴ p_{f} = \sqrt{p_{1}^{2} + p_{2}^{2}}$

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

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

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