JEE PYQ: Experimental Physics Question 21
Question 21 - 2019 (09 Apr 2019 Shift 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:
(1) $\frac{\lambda_x \lambda_y}{\lambda_x + \lambda_y}$
(2) $\frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|}$
(3) $\lambda_x - \lambda_y$
(4) $\lambda_x + \lambda_y$
Show Answer
Answer: (2)
Solution
$P \propto \frac{1}{\lambda}$. Net momentum: $\frac{1}{\lambda_x} - \frac{1}{\lambda_y} = \frac{1}{\lambda}$. $\lambda = \frac{\lambda_x \lambda_y}{|\lambda_y - \lambda_x|}$.
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