Question

A particle $\text{P}$ is formed due to a completely inelastic collision of particles $\text{x}$ and $\text{y}$ having de - Broglie wavelengths ${\lambda }_x$ and ${\lambda }_y$ respectively. If $\text{x}$ and $\text{y}$ were moving in opposite directions, then the de-Broglie wavelength of $\text{P}$ is :

• A. $\frac{{\lambda }_x{\lambda }_y}{{\lambda }_x+{\lambda }_y}$
• B. ${\lambda }_x-{\lambda }_y$
• C. ${\lambda }_x+{\lambda }_y$
• D. $\frac{{\lambda }_x{\lambda }_y}{|{\lambda }_x-{\lambda }_y|}$

Answer 1

The correct option is D $\frac{{\lambda }_x{\lambda }_y}{|{\lambda }_x-{\lambda }_y|}$

$P_x-P_y=P_p$
As, $P\propto \frac{1}{\lambda }$
$\frac{1}{{\lambda }_x}-\frac{1}{{\lambda }_y}=\frac{1}{{\lambda }_p}$
$\frac{{\lambda }_y-{\lambda }_x}{{\lambda }_x{\lambda }_y}=\frac{1}{{\lambda }_p}$
$\therefore {\lambda }_p=\frac{{\lambda }_x{\lambda }_y}{|{\lambda }_y-{\lambda }_x|}$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.