Question

Two particles move at right angles 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

The correct option is D $\frac{1}{{\lambda }^2}=\frac{1}{{\lambda }_1^2}+\frac{1}{{\lambda }_2^2}$

${\overrightarrow{P}}_1=\frac{h}{{\lambda }_1}\hat{i}$
and ${\overrightarrow{P}}_2=\frac{h}{{\lambda }_2}\hat{j}$
Using momentum conservation,
$\overrightarrow{P}={\overrightarrow{P}}_1+{\overrightarrow{P}}_2$
$=\frac{h}{{\lambda }_1}\hat{i}+\frac{h}{{\lambda }_2}\hat{j}$
$|\overrightarrow{P}|=\sqrt{{\left(\frac{h}{{\lambda }_1}\right)}^2+{\left(\frac{h}{{\lambda }_2}\right)}^2}$
$⟹\frac{h}{\lambda }=\sqrt{{\left(\frac{h}{{\lambda }_1}\right)}^2+{\left(\frac{h}{{\lambda }_2}\right)}^2}$
$⟹\frac{1}{{\lambda }^2}=\frac{1}{{\lambda }_1^2}+\frac{1}{{\lambda }_2^2}$