Ncert Exercises & Solutions

Two particles A and B of de-Broglie wavelength combine to form a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one-dimensional)

Hint: The de-Broglie wavelength depends on the momentum of the particle.

Step 1: Conserve the momentum.

Given from conservation of momentum,

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

$\begin{aligned} \Rightarrow \frac{h}{λ_{C}} & = \frac{h}{λ_{A}} + \frac{h}{λ_{B}} [∵ λ = \frac{h}{m v} = \frac{h}{p} \Rightarrow p = \frac{h}{λ}] \\ \Rightarrow \frac{h}{λ_{C}} & = \frac{h λ_{B} + h λ_{A}}{λ_{A} λ_{B}} \\ \Rightarrow \frac{λ_{C}}{h} & = \frac{λ_{A} λ_{B}}{h λ_{A} + h λ_{B}} \Rightarrow λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}} \end{aligned}$

Step 2: Find the de-Broglie wavelength in each case.

Case l: Suppose both $p_{A} and p_{B}$ are positive.

then, $λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case ll: When both $p_{A} and p_{B}$ are negative.

then, $λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case lll: When $p_{A} > 0 p_{B} < 0 i . e ., p_{A}$ is positive and $p_{B}$ is negative,

$\begin{array}{l} \frac{h}{λ_{C}} = \frac{h}{λ_{A}} - \frac{h}{λ_{B}} = \frac{(λ_{B} - λ_{A}) h}{λ_{A} λ_{B}} \\ λ_{C} = \frac{λ_{A} λ_{B}}{λ_{B} - λ_{A}} \end{array}$

Case lV: $p_{A} > 0 and p_{B} < 0 ie ., p_{A}$ is negative and $p_{B}$ is positive,

$\begin{aligned} ∴ \frac{h}{λ_{C}} & = \frac{- h}{λ_{A}} + \frac{h}{λ_{B}} \\ \Rightarrow = \frac{(λ_{A} - λ_{B}) h}{λ_{A} λ_{B}} \Rightarrow λ_{C} = \frac{λ_{A} λ_{B}}{λ_{A} - λ_{B}} \end{aligned}$

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