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.

List-I		List-II
(P)	$${}^{238}_{92}U \to {}^{234}_{91}Pa$$	(1)	one $$\alpha$$ particle and one $$\beta^+$$ particle
(Q)	$${}^{214}_{82}Pb \to {}^{210}_{82}Pb$$	(2)	three $$\beta^-$$ particles and one $$\alpha$$ particle
(R)	$${}^{210}_{81}Tl \to {}^{206}_{82}Pb$$	(3)	two $$\beta^-$$ particles and one $$\alpha$$ particle
(S)	$${}^{228}_{91}Pa \to {}^{224}_{88}Ra$$	(4)	one $$\alpha$$ particle and one $$\beta^-$$ particle
		(5)	one $$\alpha$$ particle and two $$\beta^+$$ particles

In every nuclear decay, both the mass number $$A$$ and the atomic number $$Z$$ must balance on the two sides of the reaction.
  • $$\alpha$$-particle : $$\ ^4_2He \; ( \Delta A = -4,\; \Delta Z = -2)$$
  • $$\beta^-$$-particle : $$\ ^0_{-1}e \; ( \Delta A = 0,\; \Delta Z = +1)$$
  • $$\beta^+$$-particle : $$\ ^0_{+1}e \; ( \Delta A = 0,\; \Delta Z = -1)$$

$$^{238}_{92}U \;\longrightarrow\; {}^{234}_{91}Pa$$
Required changes: $$\Delta A = 238-234 = 4,\; \Delta Z = 92-91 = 1$$ (mass ↓4, atomic ↓1).
One $$\alpha$$ gives $$\Delta A = 4, \Delta Z = 2$$ too much fall in $$Z$$. Adding one $$\beta^-$$ raises $$Z$$ by 1, giving net $$\Delta A = 4, \Delta Z = 1$$ as needed.
Emission: one $$\alpha$$ + one $$\beta^-$$ ⇒ List-II (4).

$$^{214}_{82}Pb \;\longrightarrow\; {}^{210}_{82}Pb$$
Required changes: $$\Delta A = 4,\; \Delta Z = 0$$.
One $$\alpha$$ accounts for $$\Delta A = 4, \Delta Z = 2$$. To restore $$Z$$, two $$\beta^-$$ (each +1) add $$\Delta Z = -2+2 = 0$$.
Emission: one $$\alpha$$ + two $$\beta^-$$ ⇒ List-II (3).

$$^{210}_{81}Tl \;\longrightarrow\; {}^{206}_{82}Pb$$
Required changes: $$\Delta A = 4,\; \Delta Z = -1$$ (mass ↓4, atomic ↑1).
One $$\alpha$$ gives $$\Delta A = 4,\; \Delta Z = 2$$ too low atomic number. Three $$\beta^-$$ (total +3) correct it: $$-2 + 3 = +1$$.
Emission: one $$\alpha$$ + three $$\beta^-$$ ⇒ List-II (2).

$$^{228}_{91}Pa \;\longrightarrow\; {}^{224}_{88}Ra$$
Required changes: $$\Delta A = 4,\; \Delta Z = 3$$ (mass ↓4, atomic ↓3).
One $$\alpha$$ produces $$\Delta Z = 2$$; an extra drop of 1 is still needed. One $$\beta^+$$ (−1) gives the balance: $$-2 -1 = -3$$.
Emission: one $$\alpha$$ + one $$\beta^+$$ ⇒ List-II (1).

Thus the matches are:
P → 4,  Q → 3,  R → 2,  S → 1.

Option A which is: P → 4, Q → 3, R → 2, S → 1