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 _2 of the final particle, is given by :

1 \over \lambda ^2 = 1 \over \lambda _1^2 + 1 \over \lambda _2^2

\lambda = \lambda _1 + \lambda _2 \over 2

\lambda = \sqrt \lambda _1\lambda _2

2 \over \lambda = 1 \over \lambda _1 + 1 \over \lambda _2

Dual Nature of Radiation PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q48 with Solution

View written solutionFree

Correct answer: B

For a particle,

\lambda = \frac{h}{p}

so momentum is

p = \frac{h}{\lambda}

Thus for the two particles,

p_1 = \frac{h}{\lambda_1}, \qquad p_2 = \frac{h}{\lambda_2}

The two particles move at right angles to each other, so their momenta are perpendicular.

After a perfectly inelastic collision, they stick together and move as one particle/system. Total momentum is conserved:

\vec p = \vec p_1 + \vec p_2

Since $\vec p_1 \perp \vec p_2$ ,

p = \sqrt{p_1^2 + p_2^2}

Substitute $p_1$ and $p_2$ :

p = \sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2}

p = h\sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}}

If the final de-Broglie wavelength is $\lambda$ , then

\lambda = \frac{h}{p}

So,

\lambda = \frac{h}{h\sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}}}

\lambda = \frac{1}{\sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}}}

Now squaring both sides:

\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}

This matches Option B:

\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}

Therefore, the correct answer is B.