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 :-

\lambda _x\lambda _y \over | \lambda _x - \lambda _y |

\lambda _x\lambda _y \over \lambda _x + \lambda _y

Dual Nature of Radiation PYQ — JEE Main 2019 · 9 Apr · Shift 2 · Q62 with Solution

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Correct answer: B

For any particle,

\lambda = \frac{h}{p}

so momentum is

p = \frac{h}{\lambda}

Thus for particles $x$ and $y$ ,

p_x = \frac{h}{\lambda_x}, \qquad p_y = \frac{h}{\lambda_y}

Given that $x$ and $y$ move in opposite directions, their momenta are opposite.

If the collision is completely inelastic, they stick together to form particle $P$ . By conservation of linear momentum,

p_P = |p_x - p_y|

So,

p_P = \left|\frac{h}{\lambda_x} - \frac{h}{\lambda_y}\right| = h\left|\frac{1}{\lambda_x} - \frac{1}{\lambda_y}\right|

Again using

\lambda_P = \frac{h}{p_P}

we get

\lambda_P = \frac{h}{h\left|\frac{1}{\lambda_x} - \frac{1}{\lambda_y}\right|} = \frac{1}{\left|\frac{1}{\lambda_x} - \frac{1}{\lambda_y}\right|}

Now simplify:

\lambda_P = \frac{1}{\left|\frac{\lambda_y - \lambda_x}{\lambda_x\lambda_y}\right|} = \frac{\lambda_x\lambda_y}{|\lambda_x - \lambda_y|}

This corresponds to:

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

which is Option B.

Stored correct answer: B

Our derived answer: B

So the derived answer agrees with the stored answer.