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Two particles A and B of de- Broglie wavelengths $λ_{1}$ and $λ_{2}$ respectively are combined to form a particle C. The process conserves momentum. If $λ_{2} = 2 λ_{1}$ then mark the correct options

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de- Broglie wavelength associated with particle C may be $\frac{2}{3} λ_{1}$

de- Broglie wavelength associated with the particle C may be $\frac{2}{3} λ_{2}$

de-Broglie wavelength associated with the particle C may be $2 λ_{1}$

de-Broglie wavelength associated with the particle C may be $2 λ_{2}$

answer is A, C.

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Detailed Solution

Momentum is conserved

$p_{C} = p_{A} + p_{B}$

$\Rightarrow$ $\frac{h}{λ_{C}} = \frac{h}{λ_{A}} + \frac{h}{λ_{B}}$ $(w h e n p_{A}, p_{B} > 0 o r p_{A,} p_{B} < 0)$

$\Rightarrow$ $λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

$\Rightarrow$ $λ_{C} = \frac{λ_{1} λ_{2}}{λ_{1} + λ_{2}} = \frac{λ_{1} .2 λ_{1}}{λ_{1} + 2 λ_{1}} = \frac{2 λ_{1}^{2}}{3 λ_{1}} = \frac{2}{3} λ_{1}$

When $p_{A} > 0, p_{B} < 0$

Or $p_{A} < 0, p_{B} > 0$

Then, Eq. (i) gives

$λ_{C} = \frac{λ_{B} λ_{A}}{|λ_{A} - λ_{b}|} = \frac{2 λ_{1}^{2}}{λ_{1}} = 2 λ_{1}$

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