Alpha Decay and Nuclear Structure

An alpha particle is a helium nucleus. It consists of 2 protons and 2 neutrons. Therefore, an alpha particle has a proton number (atomic number, Z) of 2 and a nucleon number (mass number, A) of 4.

When a nucleus undergoes alpha decay, it emits an alpha particle. This process changes the structure of the original nucleus (the parent nucleus) into a new nucleus (the daughter nucleus).

The changes in the proton number and nucleon number during alpha decay can be represented by the following general nuclear equation:

$$ {}_{Z}^{A}X \rightarrow {}_{Z-2}^{A-4}Y + {}_{2}^{4}\alpha $$

Where:

• $${}_{Z}^{A}X$$ is the parent nucleus with proton number Z and nucleon number A.
• $${}_{Z-2}^{A-4}Y$$ is the daughter nucleus with proton number $Z-2$ and nucleon number $A-4$.
• $${}_{2}^{4}\alpha$$ is the alpha particle.

In this question, the initial atom (parent nucleus) has:

• Proton number, $Z = 84$
• Nucleon number, $A = 216$

It emits an alpha particle, which has:

• Proton number, $Z_\alpha = 2$
• Nucleon number, $A_\alpha = 4$

To find the structure of the new nucleus (daughter nucleus), we subtract the proton number and nucleon number of the alpha particle from the original nucleus:

New Proton number ($Z'$):

$$ Z' = Z - Z_\alpha = 84 - 2 = 82 $$

New Nucleon number ($A'$):

$$ A' = A - A_\alpha = 216 - 4 = 212 $$

So, the new nucleus has a proton number of 82 and a nucleon number of 212.

Let's look at the options:

1. Proton number 82, Nucleon number 212
2. Proton number 82, Nucleon number 214
3. Proton number 85, Nucleon number 215

Our calculated values match Option 1.

Therefore, the structure of the new nucleus after the emission of an alpha particle is proton number 82 and nucleon number 212.