Question

Two particles are moving perpendicular to each other with de-Broglie wavelengths ${\lambda }_1$ and ${\lambda }_2$ if they collide and stick then the de-Broglie wavelength of the system after the collision is,

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

Answer 1

The correct option is A

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

Step 1: Given

The de-Broglie wavelengths of the given particles are ${\lambda }_1$ and ${\lambda }_2$.

Step 2: Calculating de-Broglie wavelength

Let the wavelength of the particle after collision be $\lambda$.

Let the momentum of the first and second particles be $P_1$and $P_2$ and momentum of the particle after collision be $P$.

From the conservation of momentum,

$P_1\hat{i}+P_2\hat{j}=\overrightarrow{P}[\because boththegivenparticlesaremovingperpendicular.]$

$$\begin{gathered} \Rightarrow \sqrt{{P_1}^2+{P_2}^2}=P \\ \Rightarrow \sqrt{{\left(\frac{h}{{\lambda }_1}\right)}^2+{\left(\frac{h}{{\lambda }_2}\right)}^2}=\frac{h}{\lambda }\left[Fromde-broglieequation:P=\frac{h}{\lambda }\right] \\ \Rightarrow \sqrt{\frac{1}{{{\lambda }_1}^2}+\frac{1}{{{\lambda }_2}^2}}=\frac{1}{\lambda } \\ \Rightarrow \frac{{\lambda }_1\times {\lambda }_2}{\sqrt{{{\lambda }_1}^2+{{\lambda }_2}^2}}=\lambda \end{gathered}$$

Hence, the de-Broglie wavelength of the system after the collision is $\lambda =\frac{{\lambda }_1{\lambda }_2}{\sqrt{{{\lambda }_1}^2+{{\lambda }_2}^2}}$.

Therefore, option (A) is the correct answer.