Animated Solution for Physics - Dual Nature of Matter and Radiation: 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 $5m_1$ ($6m_1=M+m_N$), respectively. If the de-Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$, then de-Broglie wavelength of the other nucleus will be

A Quiet Beginning

Imagine a heavy nucleus of mass $M$ sitting quietly in a reactor. A slowly moving neutron of mass $m_N$ approaches it. Because the neutron is moving very slowly, its momentum is practically zero ($p\approx 0$). When the nucleus absorbs this neutron, it forms a highly unstable compound nucleus. Since the initial momentum of the neutron was negligible and the original nucleus was at rest, the total initial momentum of our entire system is effectively zero.

Conservation is King

This newly formed compound nucleus cannot hold itself together and violently undergoes fission, breaking apart into two fragments. We are told the masses of these fragments are $m_1$ and $5m_1$.

Now, here is the most crucial physical principle at play: during this explosive split, there are no external forces acting on the system. The forces tearing the nucleus apart are entirely internal. Therefore, the Law of Conservation of Linear Momentum must hold perfectly true.

Since the initial momentum was zero, the vector sum of the final momenta of the two fragments must also be zero:

This simple equation tells us a profound story: ${\vec{p}}_1=-{\vec{p}}_2$. The two fragments must fly off in exactly opposite directions, and more importantly, the magnitudes of their momenta must be exactly equal ($|{\vec{p}}_1|=|{\vec{p}}_2|=p$).

Why Mass Doesn't Matter

Now, let's bring in Louis de-Broglie's brilliant hypothesis. The de-Broglie wavelength $\lambda$ associated with any moving particle is given by Planck's constant $h$ divided by the particle's momentum $p$:

Notice what is missing from this equation? Mass! The de-Broglie wavelength does not care if the particle is a tiny electron or a massive bowling ball; it only cares about the momentum. The problem deliberately gave us the mass ratio ($m_1$ and $5m_1$) as a psychological trap to make us think the wavelengths would be different.

Equal and Opposite

Because both fragments share the exact same momentum magnitude $p$, their de-Broglie wavelengths must be identical.

Therefore, the de-Broglie wavelength of the heavier nucleus is also exactly $\lambda$.

Kinetic Energy

While their momenta and wavelengths are identical, their kinetic energies are a completely different story. Kinetic energy $K$ can be expressed in terms of momentum as $K=\frac{p^2}{2m}$.

Since $p$ is the same for both, the kinetic energy is inversely proportional to the mass.

The lighter fragment flies away much faster and carries five times more kinetic energy than the heavier fragment! But in the quantum realm of matter waves, their wavelengths remain perfectly matched.