Animated Solution for Physics - Atoms and Nuclei: When ${\text{U}}^{238}$ nucleus originally at rest, decays by emitting an alpha particle having a speed $u$, the recoil speed of the residual nucleus is

Imagine a heavy, unstable nucleus resting peacefully. Suddenly, it undergoes radioactive decay, violently ejecting a smaller particle. What happens to the remaining chunk of the nucleus? It doesn't just sit there; it recoils backward, much like a cannon kicking back when it fires a cannonball. This phenomenon is a classic demonstration of the Conservation of Linear Momentum.

In our specific problem, we start with a Uranium-238 ($U^{238}$) nucleus that is initially at rest. Because it is stationary, its initial momentum is exactly zero.

When it decays, it emits an alpha particle. An alpha particle is essentially a Helium nucleus, which has a mass number of $4$. The problem states that this alpha particle is ejected with a speed $u$.

What about the residual nucleus? Since the original mass number was $238$ and it lost a mass of $4$, the remaining nucleus (let's call it $X$) will have a mass number of $238-4=234$. Let's assume this residual nucleus recoils with a velocity $v$ in the opposite direction.

Since the decay is an internal process and no external forces are acting on the system, the total linear momentum must be conserved. The final momentum of the system must equal the initial momentum.

The final momentum is the vector sum of the momentum of the alpha particle and the momentum of the residual nucleus. Taking the direction of the alpha particle as positive, the velocity of the recoiling nucleus will be $-v$.

Substituting the mass numbers for the masses (since the actual mass is roughly proportional to the mass number):

Now, we simply solve this algebraic equation for the recoil speed $v$:

This gives us the magnitude of the recoil velocity. You might wonder why we don't include a negative sign in our final answer. The term recoil inherently signifies the backward direction. When asked for the "recoil speed," we are only interested in the magnitude of that backward velocity. Therefore, the correct expression for the recoil speed is $\frac{4u}{234}$.