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(B) Let the masses of the two particles be $m$ and their speeds be $v$. The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$,so $mv = \frac{

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$2\ \lambda$

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(B) Let the masses of the two particles be $m$ and their speeds be $v$. The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$,so $mv = \frac{h}{\lambda}$.In the centre of mass $(CM)$ frame,the velocity of particle $1$ is $\vec{v}_{1cm} = \vec{v}_1 - \vec{v}_{cm}$.The velocity of the centre of mass is $\vec{v}_{cm} = \frac{m\vec{v}_1 + m\vec{v}_2}{2m} = \frac{\vec{v}_1 + \vec{v}_2}{2}$.Thus,$\vec{v}_{1cm} = \vec{v}_1 - \frac{\vec{v}_1 + \vec{v}_2}{2} = \frac{\vec{v}_1 - \vec{v}_2}{2}$.The magnitude of the relative velocity is $|\vec{v}_{1cm}| = \frac{1}{2} |\vec{v}_1 - \vec{v}_2|$.Using the law of cosines for the vector subtraction of two vectors of equal magnitude $v$ at an angle $	heta = 60^o$:$|\vec{v}_1 - \vec{v}_2| = \sqrt{v^2 + v^2 - 2v^2 \cos(60^o)} = \sqrt{2v^2 - 2v^2(0.5)} = \sqrt{v^2} = v$.Therefore,$|\vec{v}_{1cm}| = \frac{v}{2}$.The new de-Broglie wavelength is $\lambda' = \frac{h}{m|\vec{v}_{1cm}|} = \frac{h}{m(v/2)} = 2 \frac{h}{mv} = 2\lambda$.

Two particles of equal masses are moving with equal speeds at an angle $60^o$. The de-Broglie wavelength of these particles is $\lambda$. Find the de-Broglie wavelength of the particles in the frame of the centre of mass of the particles.

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