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(D) Let the momenta of the two particles be $\vec{P}_1$ and $\vec{P}_2$. Given they move at right angles,we can represent them as $\vec{P}_1 = \frac{h

Vedclass · Accepted Answer

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

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(D) Let the momenta of the two particles be $\vec{P}_1$ and $\vec{P}_2$. Given they move at right angles,we can represent them as $\vec{P}_1 = \frac{h}{\lambda_1} \hat{i}$ and $\vec{P}_2 = \frac{h}{\lambda_2} \hat{j}$.Using the law of conservation of linear momentum,the momentum of the final particle $\vec{P}$ is the vector sum of the initial momenta:$\vec{P} = \vec{P}_1 + \vec{P}_2 = \frac{h}{\lambda_1} \hat{i} + \frac{h}{\lambda_2} \hat{j}$.The magnitude of the final momentum is:$|\vec{P}| = \sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2}$.Since the de Broglie wavelength of the final particle is $\lambda = \frac{h}{|\vec{P}|}$,we have:$\frac{h}{\lambda} = \sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2}$.Squaring both sides and dividing by $h^2$ gives:$\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}$.

Two particles move at right angles to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles undergo a perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle is given by

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