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A particle moving in the positive X direction has de Broglie wavelength $λ_{1}$ . Another particle also moving in the same direction has de Broglie wavelength $λ_{2}$ . The two particles make a collision and after the collision they move together as a single particle. The de Broglie wavelength of this single particle is given by

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$\frac{λ_{1} + λ_{2}}{2}$

$\sqrt{λ_{1} λ_{2}}$

$\frac{2 λ_{1} λ_{2}}{λ_{1} + λ_{2}}$

$\frac{λ_{1} λ_{2}}{λ_{1} + λ_{2}}$

answer is D.

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Detailed Solution

$λ_{1} = \frac{h}{p_{1}} \Rightarrow p_{1} = \frac{h}{λ_{1}}$

$similarly, p_{2} = \frac{h}{λ_{2}}$

$after collision, p = p_{1} + p_{2}$

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

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

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