JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Conservation of Momentum

The de Broglie wavelength is tied to momentum by $\lambda =\frac{h}{p}$. In a completely inelastic head-on collision the particles stick together, so the final momentum is the vector sum of the two; moving in opposite directions, their momenta subtract: $p=p_x-p_y$, that is $\frac{h}{\lambda }=\frac{h}{{\lambda }_x}-\frac{h}{{\lambda }_y}$. Dividing out $h$, $\frac{1}{\lambda }=\frac{1}{{\lambda }_x}-\frac{1}{{\lambda }_y}=\frac{{\lambda }_y-{\lambda }_x}{{\lambda }_x{\lambda }_y}$, so taking magnitudes $\lambda =\frac{{\lambda }_x{\lambda }_y}{|{\lambda }_x-{\lambda }_y|}$.