Animated Solution for Physics - Atoms and Nuclei: List-I shows different radioactive decay processes and List-II provides possible emitted particles. Match each entry in List-I with an appropriate entry from List-II, and choose the correct option.

Welcome to the fascinating world of radioactive decay, where elements literally transform into entirely different elements! This isn't medieval alchemy; it's the rigorous, mathematically beautiful reality of nuclear physics. In this problem, we are presented with four distinct nuclear reactions. Our mission is to act as nuclear detectives and deduce exactly which particles were emitted during these transformations.

To solve this, we don't need complex quantum mechanics. We just need to rely on two unbreakable laws of the universe: the conservation of mass number and the conservation of atomic number (charge).

Every nuclear reaction can be written in a general form:

1. Conservation of Mass Number ($A$): The total mass number before the decay must equal the total mass number after the decay. Since $\beta$ particles (electrons or positrons) have a mass number of zero, any change in the mass number is entirely due to the emission of $\alpha$ particles (Helium nuclei). Each $\alpha$ particle carries away a mass number of $4$. Therefore, the number of $\alpha$ particles emitted is simply:

2. Conservation of Atomic Number ($Z$): The total charge must also be conserved. The original atomic number $Z$ must equal the new atomic number $Z^{\prime }$, plus $2$ times the number of $\alpha$ particles, plus the net charge carried away by the $\beta$ particles.

If $\Delta Z_{\beta }$ is negative, it means ${\beta }^{-}$ particles (electrons, charge $-1$) were emitted. If $\Delta Z_{\beta }$ is positive, it means ${\beta }^{+}$ particles (positrons, charge $+1$) were emitted.

Step-by-Step

Let's apply our detective tools to each reaction.

Reaction P: ${}_{92}^{238}\text{U}\to {}_{91}^{234}\text{Pa}$ First, we check the mass number. It drops from $238$ to $234$, a difference of $4$. This immediately tells us exactly $1$ $\alpha$ particle is emitted. Now for the charge: $92=91+2(1)+\Delta Z_{\beta }$. Solving this gives $\Delta Z_{\beta }=-1$. Conclusion: $1\alpha$ and $1{\beta }^{-}$ are emitted. This matches option (4).

Reaction Q: ${}_{82}^{214}\text{Pb}\to {}_{82}^{210}\text{Pb}$ The mass number decreases from $214$ to $210$, a difference of $4$, so we have $1$ $\alpha$ particle. For the atomic number: $82=82+2(1)+\Delta Z_{\beta }$. This means $\Delta Z_{\beta }=-2$ to balance the equation. Conclusion: $1\alpha$ and $2{\beta }^{-}$ are emitted. This matches option (3).

Reaction R: ${}_{81}^{210}\text{Tl}\to {}_{82}^{206}\text{Pb}$ The mass number difference is $210-206=4$, yielding $1$ $\alpha$ particle. The atomic number equation is $81=82+2(1)+\Delta Z_{\beta }$. Solving this gives $\Delta Z_{\beta }=-3$. Conclusion: $1\alpha$ and $3{\beta }^{-}$ are emitted. This matches option (2).

Reaction S: ${}_{91}^{228}\text{Pa}\to {}_{88}^{224}\text{Ra}$ The mass number drops by $4$, so $1$ $\alpha$ particle is emitted. The atomic number equation is $91=88+2(1)+\Delta Z_{\beta }$. This time, $\Delta Z_{\beta }=+1$! This is a crucial difference. It means a positive charge was emitted. Conclusion: $1\alpha$ and $1{\beta }^{+}$ (positron) are emitted. This matches option (1).

By systematically applying the conservation laws, we have successfully decoded the entire matrix. The beauty of physics lies in its consistency; no matter how complex the nucleus, the fundamental rules of arithmetic always hold true.