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(C) According to the principle of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collisio

Vedclass · Accepted Answer

$\frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|}$

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(C) According to the principle of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision.Let the momenta of particles $x$ and $y$ be $\vec{p}_x$ and $\vec{p}_y$ respectively.Since they move in opposite directions,we take one direction as positive and the other as negative.The magnitude of momentum is related to the de-Broglie wavelength by $p = \frac{h}{\lambda}$.Let $p_x = \frac{h}{\lambda_x}$ and $p_y = \frac{h}{\lambda_y}$.The final momentum of particle $P$ is $p_f = |p_x - p_y|$.Substituting the values,$p_f = |\frac{h}{\lambda_x} - \frac{h}{\lambda_y}| = h |\frac{1}{\lambda_x} - \frac{1}{\lambda_y}|$.Since $p_f = \frac{h}{\lambda}$,we have $\frac{h}{\lambda} = h |\frac{\lambda_y - \lambda_x}{\lambda_x \lambda_y}|$.Therefore,$\lambda = \frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|}$.

$A$ 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

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