Question:
Aparticle ' $\mathrm{P}$ ' is formed due to a completely inelastic collision of particles ' $x$ ' and ' $y$ ' having de-Broglie wavelengths ' $\gamma_{x}$ ' and ' $\gamma_{y}$ ' respectively. If $x$ and $y$ were moving in opposite directions, then the de-Broglie wavelength of ' $\mathrm{P}$ ' is:
Correct Option: 2
Solution:
(2) $P_{1}-P_{2}=\left(P_{1}+P_{2}\right)=P$
As $P \propto \frac{1}{\lambda}$
or $\frac{1}{\lambda_{x}}-\frac{1}{\lambda_{y}}=\frac{1}{\lambda}$
or $\frac{\lambda_{y}-\lambda_{x}}{\lambda_{x} \lambda_{y}}=\frac{1}{\lambda}$
$\therefore \lambda=\frac{\lambda_{x} \lambda_{y}}{\mid \lambda_{y}-\lambda_{x}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.