One kind of radioactivity is called "alpha decay." For example, the nucleus of a radium-220 atom can spontaneously split into a radon-216 nucleus plus an alpha particle (a helium nucleus containing two protons and two neutrons). Consider a radium-220 nucleus that is initially at rest. It spontaneously decays, and the alpha particle travels off in the $+z$ direction. What can you conclude about the motion of the new radon-216 nucleus? Be as precise as you can, and explain your reasoning.

• The radium-220 nucleus split into radon-216 and an alpha particle.
• The radium-220 nucleus is initially at rest.

The total momentum of the system before the split is equal to the total momentum of the system after the split. Which representing as

$P_i=P_f$

Where$P_i$ is the initial momentum, and$P_f$ is the final momentum.

Here radium-220 is initially at rest and split into an alpha particle and radon-216.

So, we have that the overall momentum of the system must be considered.

So we have the initial momentum of the system is zero.

$P_i=0$

Which means

$$\begin{gathered} P_a+P_{216}=0 \\ {P}_a=-P_{216} \end{gathered}$$

As$P_a$ is in the z-direction, then radon-216 will be in the z-direction.So radon-216 travels off in the opposite direction to the alpha particle.

Thus, the motion of the new radon-216 nucleus is –z-direction.