Two particles $A$ and $B$ of de Broglie wavelengths $\lambda_A$ and $\lambda_B$ combine to form a pa

Question

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

Professor PC Thomas Classes · Answer

(d) : For one dimensional motion, $ \overrightarrow{p_C}=\overrightarrow{p_A}+\overrightarrow{p_B} $ If $p_A, p_B>0$ or $p_A, p_B<0$, i.e., $p_A$ \ and $p_B$ are in same direction. $ p_C=p_A+p_B $ $\frac{h}{\lambda_C}=\frac{h}{\lambda_A}+\frac{h}{\lambda_B}=h\left(\frac{\lambda_A+\lambda_B}{\lambda_A \lambda_B} ight)$ $\Rightarrow \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$ If $p_A>0, p_B<0$ or $p_A<0, p_B>0$ ( $p_A$ and $p_B$ are in opposite direction) $\begin{aligned} & p_C=\left|p_A-p_B ight| \ & \frac{h}{\lambda_C}=\left|\frac{h}{\lambda_A}-\frac{h}{\lambda_B} ight|=\frac{h\left|\lambda_A-\lambda_B ight|}{\lambda_A \lambda_B} ;\end{aligned}$ $\lambda_C=\frac{\lambda_A \lambda_B}{\left|\lambda_A-\lambda_B ight|}$