JEE PYQ: Experimental Physics Question 17
Question 17 - 2019 (08 Apr 2019 Shift 1)
Two particles move at right angle to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle, is given by:
(1) $\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}$
(2) $\lambda = \sqrt{\lambda_1 \lambda_2}$
(3) $\lambda = \frac{\lambda_2 + \lambda_2}{2}$
(4) $\frac{2}{\lambda} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$
Show Answer
Answer: (1)
Solution
$p_1 = \frac{h}{\lambda_1}$, $p_2 = \frac{h}{\lambda_2}$. Since particles move at right angles: $p_f = \sqrt{p_1^2 + p_2^2}$. $\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}$.
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