Question

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 :

• A. $\lambda = \frac{\lambda_{1} + \lambda_{2}}{2} $
• B. $\frac{2}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}} $
• C. $\lambda = \sqrt{ \lambda_1 \lambda_2} $
• D. $\frac{1}{\lambda^2} = \frac{1}{\lambda_{1}^{2}} + \frac{1}{\lambda_{2}^{2}} $

Answer 1

Answer: (D)

$\vec{P}_{2}=\frac{h}{\lambda_{1} } \hat{i} $
& $ \vec{P}_{2} = \frac{h}{\lambda_{2}} \hat{j} $
Using momentum conservation
$ \vec{P} = \vec{P}_{1} + \vec{P}_{2} $
$ = \frac{h}{\lambda_{1}} \hat{i} + \frac{h}{\lambda_{2}} \hat{j} $
$ \left|\vec{P}\right| = \sqrt{\left(\frac{h}{\lambda_{1}}\right)^{2} + \left(\frac{h}{\lambda_{2}}\right)^{2}} $
$ \frac{h}{\lambda} = \sqrt{\left(\frac{h}{\lambda_{1}}\right)^{2} +\left(\frac{h}{\lambda_{2}}\right)^{2}} $
$ \frac{1}{\lambda^{2}} = \frac{1}{\lambda_{1}^{2}} + \frac{1}{\lambda_{2}^{2}} $