A particle ' P ' is formed due to a completely inelastic collision

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:

  1. (1) $\frac{\gamma_{x} \gamma_{y}}{\gamma_{x}+\gamma_{y}}$

  2. (2) $\frac{\gamma_{x} \gamma_{y}}{\left|\gamma_{x}-\gamma_{y}\right|}$

  3. (3) $\gamma_{x}-\gamma_{y}$

  4. (4) $\gamma_{x}+\gamma_{y}$


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 $\frac{\lambda_{y}-\lambda_{x}}{\lambda_{x} \lambda_{y}}=\frac{1}{\lambda}$

$\therefore \lambda=\frac{\lambda_{x} \lambda_{y}}{\mid \lambda_{y}-\lambda_{x}}$

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