Q 12-11-162JEE MainJEE Main 2019 (8 Apr, Shift 1)Easy
Two particles move at right angle to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle, is given by
Answer: (D) $\dfrac1{\lambda^2} = \dfrac1{\lambda_1^2} + \dfrac1{\lambda_2^2}$
Momenta $p_1 = h/\lambda_1$ and $p_2 = h/\lambda_2$ are perpendicular. Momentum is conserved, so the combined particle has
$$p = \sqrt{p_1^2 + p_2^2} \Rightarrow \frac{h^2}{\lambda^2} = \frac{h^2}{\lambda_1^2} + \frac{h^2}{\lambda_2^2}$$
$$\frac1{\lambda^2} = \frac1{\lambda_1^2} + \frac1{\lambda_2^2}$$
Solution by Sreeraj P, M.Sc Physics