Question

Two particles moving at right angles have their de-Broglie wavelengths ${\lambda }_1$ and ${\lambda }_2$. If their collision is perfectly inelastic, what is the final wavelength $\lambda$, in terms of ${\lambda }_1$ and ${\lambda }_2?$

• A. $\lambda ={\lambda }_1+{\lambda }_2$
• B. $\frac{1}{\lambda }=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$
• C. ${\lambda }^2={\lambda }_1^2+{\lambda }_2^2$
• D. $\frac{1}{{\lambda }^2}=\frac{1}{{\lambda }_1^2}+\frac{1}{{\lambda }_2^2}$

Answer 1

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

As momentum is conserved, $\overrightarrow{p}=\overrightarrow{p_1}+\overrightarrow{p_2}$
Since, the particles are at right angles,
$\Rightarrow p^2=p_1^2+p_2^2$

As, $p=\frac{h}{\lambda }\Rightarrow \frac{h^2}{{\lambda }^2}=\frac{h^2}{{\lambda }_1^2}+\frac{h^2}{{\lambda }_2^2}$

$\therefore \frac{1}{{\lambda }^2}=\frac{1}{{\lambda }_1^2}+\frac{1}{{\lambda }_2^2}$

Hence, $\left(D\right)$ is the correct answer.

Key concept: For a perfectly inelastic collision, only momentum is conserved and the objects stick together after collision. $\begin{array}{c}\text{Why this question?}\\ \text{Application of the concept of de-Broglie wavelength and momentum conservation.}\end{array}$