Question

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.

Short answer

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

Step-by-step solution

Given Information 

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

Significance of Law of conservation of momentum

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.

Motion of the new radon-216 nucleus

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.