Question
Two particles $A$ and $B$ of de Broglie wavelengths $\lambda _1$ and $\lambda _2$ combine to form a particle $C$. The process conserves momentum. Find the de Broglie wavelength of the particle $C. \ ($The motion is one dimensional$)$.
✓
Answer
By the law of conservation of momentum,
$|p_C| = |p_A| + |p_B|$
Let us first take the case $I$ when both $p_A$ and $p_B$ are positive
then $\lambda_\text{C}=\frac{\lambda_\text{A}\lambda_\text{B}}{\lambda_\text{A}+\lambda_\text{B}}$
In case $III $ when $p_A > 0, p_B < 0$
i.e., $p_A$ is positive and $p_B$ is negative,
$\frac{\text{h}}{\lambda_\text{C}}=\frac{\text{h}}{\lambda_\text{A}}-\frac{\text{h}}{\lambda_\text{B}}=\frac{(\lambda_\text{B}-\lambda_\text{A})\text{h}}{\lambda_\text{A}\lambda_\text{B}}$
$\Rightarrow\ \lambda_\text{C}=\frac{\lambda_\text{A}\lambda_\text{B}}{\lambda_\text{B}-\lambda_\text{A}}$
And in case $IV$ when $p_A < 0, p_B > 0,$
i.e., $p_A$ is negative and $p_B$ is positive.
$\therefore\ \frac{\text{h}}{\lambda_\text{C}}=\frac{-\text{h}}{\lambda_\text{A}}+\frac{\text{h}}{\lambda_\text{B}}$
$\Rightarrow\ \frac{(\lambda_\text{A}-\lambda_\text{B})\text{h}}{\lambda_\text{A}\lambda_\text{B}}$
$\Rightarrow\ \lambda_\text{C}=\frac{\lambda_\text{A}\lambda_\text{B}}{\lambda_\text{A}-\lambda_\text{B}}$
$|p_C| = |p_A| + |p_B|$
Let us first take the case $I$ when both $p_A$ and $p_B$ are positive
then $\lambda_\text{C}=\frac{\lambda_\text{A}\lambda_\text{B}}{\lambda_\text{A}+\lambda_\text{B}}$
In case $III $ when $p_A > 0, p_B < 0$
i.e., $p_A$ is positive and $p_B$ is negative,
$\frac{\text{h}}{\lambda_\text{C}}=\frac{\text{h}}{\lambda_\text{A}}-\frac{\text{h}}{\lambda_\text{B}}=\frac{(\lambda_\text{B}-\lambda_\text{A})\text{h}}{\lambda_\text{A}\lambda_\text{B}}$
$\Rightarrow\ \lambda_\text{C}=\frac{\lambda_\text{A}\lambda_\text{B}}{\lambda_\text{B}-\lambda_\text{A}}$
And in case $IV$ when $p_A < 0, p_B > 0,$
i.e., $p_A$ is negative and $p_B$ is positive.
$\therefore\ \frac{\text{h}}{\lambda_\text{C}}=\frac{-\text{h}}{\lambda_\text{A}}+\frac{\text{h}}{\lambda_\text{B}}$
$\Rightarrow\ \frac{(\lambda_\text{A}-\lambda_\text{B})\text{h}}{\lambda_\text{A}\lambda_\text{B}}$
$\Rightarrow\ \lambda_\text{C}=\frac{\lambda_\text{A}\lambda_\text{B}}{\lambda_\text{A}-\lambda_\text{B}}$
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