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

The correct option is P → 4, Q → 3, R → 2, S → 1.

To determine the emitted particles in each radioactive decay process, we apply the conservation of mass number (A) and atomic number (Z):

1. Emitting an $\alpha$ particle $\left({\mathrm{He}}_2^4\right)$ decreases the mass number by 4 and the atomic number by 2.
2. Emitting a ${\beta }^{-}$ particle $\left(e_{-1}^0\right)$ leaves the mass number unchanged and increases the atomic number by 1.
3. Emitting a ${\beta }^{+}$ particle $\left(e_{+1}^0\right)$ leaves the mass number unchanged and decreases the atomic number by 1.

Let us analyze each process step-by-step:

(P) $U_{92}^{238}\to {\mathrm{Pa}}_{91}^{234}$

Change in mass number (A): 238 - 234 = 4.
This indicates the emission of 1 $\alpha$ particle (which reduces A by 4 and Z by 2).
After emitting 1 $\alpha$ particle, the atomic number becomes: 92 - 2 = 90.
The final product has an atomic number of 91, which requires an increase of 1 in Z.
This increase is achieved by emitting 1 ${\beta }^{-}$ particle.
Hence, process (P) emits one $\alpha$ particle and one ${\beta }^{-}$ particle.
Thus, P → 4.

(Q) ${\mathrm{Pb}}_{82}^{214}\to {\mathrm{Pb}}_{82}^{210}$

Change in mass number (A): 214 - 210 = 4.
This corresponds to the emission of 1 $\alpha$ particle.
After emitting 1 $\alpha$ particle, the atomic number becomes: 82 - 2 = 80.
The final product has an atomic number of 82, requiring an increase of 2 in Z.
This increase is achieved by emitting 2 ${\beta }^{-}$ particles.
Hence, process (Q) emits two ${\beta }^{-}$ particles and one $\alpha$ particle.
Thus, Q → 3.

(R) ${\mathrm{Tl}}_{81}^{210}\to {\mathrm{Pb}}_{82}^{206}$

Change in mass number (A): 210 - 206 = 4.
This corresponds to the emission of 1 $\alpha$ particle.
After emitting 1 $\alpha$ particle, the atomic number becomes: 81 - 2 = 79.
The final product has an atomic number of 82, requiring an increase of 3 in Z.
This increase is achieved by emitting 3 ${\beta }^{-}$ particles.
Hence, process (R) emits three ${\beta }^{-}$ particles and one $\alpha$ particle.
Thus, R → 2.

(S) ${\mathrm{Pa}}_{91}^{228}\to {\mathrm{Ra}}_{88}^{224}$

Change in mass number (A): 228 - 224 = 4.
This corresponds to the emission of 1 $\alpha$ particle.
After emitting 1 $\alpha$ particle, the atomic number becomes: 91 - 2 = 89.
The final product has an atomic number of 88, requiring a decrease of 1 in Z.
This decrease is achieved by emitting 1 ${\beta }^{+}$ particle.
Hence, process (S) emits one $\alpha$ particle and one ${\beta }^{+}$ particle.
Thus, S → 1.

Combining all the matches gives:
P → 4, Q → 3, R → 2, S → 1