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(D) According to the law of conservation of linear momentum,the total momentum of the system must remain constant.Since the initial momentum of the sy

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$\lambda$

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(D) According to the law of conservation of linear momentum,the total momentum of the system must remain constant.Since the initial momentum of the system is approximately zero (as the neutron is moving slowly),the final momentum of the two resulting nuclei must also be zero.Let $p_1$ be the momentum of the nucleus with mass $m_1$ and $p_2$ be the momentum of the nucleus with mass $5m_1$.Therefore,$p_1 + p_2 = 0$,which implies $p_1 = -p_2$,or $|p_1| = |p_2|$.The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the magnitude of the momentum.Since both nuclei have the same magnitude of momentum $(|p_1| = |p_2|)$,their de-Broglie wavelengths must also be equal.Thus,the de-Broglie wavelength of the second nucleus is also $\lambda$.

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

After absorbing a slowly moving neutron of mass $m_N$ (momentum $\sim 0$),a nucleus of mass $M$ breaks into two nuclei of masses $m_1$ and $5m_1$ $(6m_1 = M + m_N)$. If the de-Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then the de-Broglie wavelength of the other nucleus will be:

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