Modern Physics 4 Question 2
2. A particle $P$ is formed due to a completely inelastic collision of particles $x$ and $y$ having de-Broglie wavelengths $\lambda _x$ and $\lambda _y$, respectively. If $x$ and $y$ were moving in opposite directions, then the de-Broglie wavelength of $P$ is (Main 2019, 9 April II)
(a) $\lambda _x-\lambda _y$
(b) $\frac{\lambda _x \lambda _y}{\lambda _x-\lambda _y}$
(c) $\frac{\lambda _x \lambda _y}{\lambda _x+\lambda _y}$
(d) $\lambda _x+\lambda _y$
Show Answer
Solution:
- Initially,
$$ \stackrel{x}{p _x} \stackrel{p _y}{\longleftrightarrow} \stackrel{y}{\longleftarrow} $$
We have, de-Broglie wavelengths associated with particles are
$$ \begin{aligned} \lambda _x & =\frac{h}{p _x} \text { and } \lambda _y=\frac{h}{p _y} \\ \Rightarrow \quad p _x & =\frac{h}{\lambda _x} \text { and } p _y=\frac{h}{\lambda _y} \end{aligned} $$
Finally, particles collided to form a single particle.
As we know that linear momentum is conserved in collision, so
$$ \mathbf{p} _p=\left|\mathbf{p} _x-\mathbf{p} _y\right| \Rightarrow \mathbf{p} _p=\left|\frac{h}{\lambda _x}-\frac{h}{\lambda _y}\right| $$
So, de-Broglie wavelength of combined particle is
$\lambda _p=\frac{h}{\left|\mathbf{p} _p\right|}=\frac{h}{\left|\frac{h}{\lambda _x}-\frac{h}{\lambda _y}\right|}=\frac{h}{\left|\frac{h \lambda _y-h \lambda _x}{\lambda _x \lambda _y}\right|}=\frac{\lambda _x \lambda _y}{\left|\lambda _x-\lambda _y\right|}$
