Animated Solution for Physics - Atoms and Nuclei: A nucleus with $Z=92$ emits the following in a sequence $\alpha ,\alpha ,{\beta }^{-},{\beta }^{-},\alpha ,\alpha ,\alpha ,\alpha ;{\beta }^{-},{\beta }^{-},\alpha ,{\beta }^{+},{\beta }^{+},\alpha$. The $Z$ of the resulting nucleus is

This nucleus is about to undergo a long and chaotic series of radioactive decays. Our mission is to track the changes in its atomic number $Z$ as it spits out various particles one by one.

The sequence of emitted particles is given as: $\alpha ,\alpha ,{\beta }^{-},{\beta }^{-},\alpha ,\alpha ,\alpha ,\alpha ;{\beta }^{-},{\beta }^{-},\alpha ,{\beta }^{+},{\beta }^{+},\alpha$.

Instead of calculating the change step-by-step, which can be tedious and prone to silly mistakes, we can simply count the total number of each type of particle emitted. The order of emission does not matter because addition and subtraction are commutative.

Before we start counting, let's quickly review how each type of decay affects the atomic number $Z$:

1. Alpha ($\alpha$) Decay: An alpha particle is essentially a helium nucleus, denoted as ${}_2^4\text{He}$. When a nucleus emits an alpha particle, it loses 2 protons. Therefore, the atomic number decreases by 2 ($\Delta Z=-2$).

2. Beta Minus (${\beta }^{-}$) Decay: In this process, a neutron transforms into a proton while emitting an electron (${}_{-1}^0\text{e}$). Because a new proton is created, the atomic number increases by 1 ($\Delta Z=+1$).

3. Beta Plus (${\beta }^{+}$) Decay: Here, a proton transforms into a neutron while emitting a positron (${}_{+1}^0\text{e}$). Since a proton is lost, the atomic number decreases by 1 ($\Delta Z=-1$).

Now, let's scan the given sequence and tally up the particles:

- Alpha particles ($\alpha$): Counting them all up, we find a total of $8$ alpha particles. The total change in $Z$ due to alpha emissions is $8\times (-2)=-16$.

- Beta Minus particles (${\beta }^{-}$): We can spot $4$ beta minus particles in the sequence. The total change in $Z$ due to beta minus emissions is $4\times (+1)=+4$.

- Beta Plus particles (${\beta }^{+}$): There are $2$ beta plus particles near the end of the sequence. The total change in $Z$ due to beta plus emissions is $2\times (-1)=-2$.

To find the final atomic number, we simply apply these net changes to our initial $Z$:

Let's do the math carefully: $92-16=76$. Then, $76+4=80$. Finally, $80-2=78$.

The atomic number of the resulting nucleus is 78. This perfectly matches option (b).