After Absorbing a slowly moving neutron of mass $m_N$ (momentum~0) a nucleus of mass M beaks into two nuclei of masses $m_1$ and $5m_1$ ($6m_1 = M + m_N$), respectively. If the de Broglie wavelength of the nucleus with mass $m_1$ is $λ$, then de-Broglie wavelength of the other nucleus will be

$λ$

$\text{Since initial momentum of the system is } P_i = 0$

$\text{Hence final momentum is also } P_f = 0$

$\text{Let final momentum of two nuclei are } P_1\, \text{and}\, P_2 \,\text{respectively.}$

$\text{From conservation of momentum } P_1 = P_2$

$ Since \lambda = \frac{h}{p}$

$\Rightarrow \lambda_1 = \lambda_2$