Animated Solution for Physics - Atoms and Nuclei: Comprehension Passage

The Anatomy of a Decay

Imagine a heavy, unstable nucleus floating in the void, perfectly at rest. This is our parent nucleus, and it has a mass of $M+\Delta m$. Because it is at rest, its initial velocity is zero, which means its initial linear momentum is also exactly zero.

Suddenly, the nucleus undergoes spontaneous fission, splitting into two identical daughter nuclei. The problem tells us that each of these daughter nuclei has a mass of $M/2$. This is a classic scenario in nuclear physics, and to solve it, we need to rely on two of the most fundamental laws of the universe: the conservation of momentum and the conservation of energy.

Conservation of Momentum

Since there are no external forces acting on this isolated system, the total linear momentum must remain constant before and after the decay.

Mathematically, we write this as:

Because the parent nucleus was at rest, $p_{\text{initial}}=0$. Therefore, the sum of the momenta of the two daughter nuclei must also be zero:

Let's assume the first daughter nucleus moves to the right with a velocity $v_1$, and the second moves to the left with a velocity $v_2$. Substituting their masses, we get:

Solving this simple equation reveals a beautiful symmetry:

Because the two fragments have identical masses, they must fly apart in exactly opposite directions with the exact same speed to keep the universe's momentum ledger perfectly balanced at zero.

The Source of Energy

Now, a critical question arises: If the parent nucleus was at rest, where did the daughter nuclei get the kinetic energy to fly apart at such high speeds?

To answer this, we must look at the masses. The initial mass was $M+\Delta m$. The total final mass is the sum of the two daughter nuclei: $\frac{M}{2}+\frac{M}{2}=M$.

Notice something profound? A small amount of mass, exactly $\Delta m$, has vanished! This missing mass is known as the mass defect. According to Albert Einstein's legendary mass-energy equivalence principle, this missing mass hasn't truly disappeared; it has been converted into pure energy.

The total energy released ($Q$) is given by:

This released energy, $\Delta m{c}^2$, doesn't just vanish into the ether. It manifests entirely as the kinetic energy of the two flying daughter nuclei.

We can set up our master energy equation by equating the released energy to the sum of their kinetic energies:

Substituting the classical kinetic energy formula ($K=\frac{1}{2}m{v}^2$) for both fragments:

Let's simplify the right side of the equation. Half of $\frac{M}{2}{v}^2$ plus another half of $\frac{M}{2}{v}^2$ simply gives us the full $\frac{M}{2}{v}^2$:

Now, we are just one algebraic step away from our goal. We need to isolate $v$, the speed of the daughter nuclei. First, multiply both sides by 2 and divide by $M$:

Finally, take the square root of both sides:

And there we have it! The speed of each daughter nucleus is elegantly expressed in terms of the speed of light, the mass defect, and the final mass. This result perfectly matches option (b).