A nucleus of mass M,moving slowly,absorbs a n | vedclass

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(D) The initial momentum of the system is approximately zero because the nucleus is moving slowly and the neutron is absorbed. By the law of conservat

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$\lambda$

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(D) The initial momentum of the system is approximately zero because the nucleus is moving slowly and the neutron is absorbed. By the law of conservation of linear momentum,the total momentum of the two product nuclei must be equal and opposite. Let $p_1$ and $p_2$ be the magnitudes of the momenta of the two nuclei. Since the total momentum is zero,$p_1 = p_2 = p$. The de Broglie wavelength is given by $\lambda = h/p$. Since both nuclei have the same magnitude of momentum $p$,their de Broglie wavelengths will be equal. Therefore,the de Broglie wavelength of the second nucleus is also $\lambda$.

$A$ nucleus of mass $M$,moving slowly,absorbs a neutron of mass $m_N$ and then breaks into two nuclei of masses $m_1$ and $5m_1$. If the de Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then what will be the de Broglie wavelength of the other nucleus?

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