Question

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

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

Answer 1

Answer: (C)

By momentum conservation
$P_x-P_y=P_P$
$\frac{h}{{\lambda }_x}-\frac{h}{{\lambda }_y}=\frac{h}{{\lambda }_p}$
${\lambda }_p=\frac{{\lambda }_x{\lambda }_y}{|{\lambda }_y-{\lambda }_x|}$