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
$\vec{P}={\vec{P}}_1+{\vec{P}}_2$
$=\frac{h}{{\lambda }_1}\hat{i}+\frac{h}{{\lambda }_2}\hat{j}$
$\left|\vec{P}\right|=\sqrt{{\left(\frac{h}{{\lambda }_1}\right)}^2+{\left(\frac{h}{{\lambda }_2}\right)}^2}$
$\frac{h}{\lambda }=\sqrt{{\left(\frac{h}{{\lambda }_1}\right)}^2+{\left(\frac{h}{{\lambda }_2}\right)}^2}$
$\frac{1}{{\lambda }^2}=\frac{1}{{\lambda }_1^2}+\frac{1}{{\lambda }_2^2}$