The emission of an `alpha or a beta` particle by a radioactive element forms a new element. However, successive emission of some `alpha or beta`- particles may give rise to an isotope or an isobar of the original element. In many cases, positron emission or `K -` electron capture takes place, leading again to the formuation of new elements, alongwith the emission of neutrinos or antineutrinos. These emission also change the neutron/proton `(n//p)` ratio so that they give rise to stable isotopes which lie in the stability belt. However, in any disintegration reaction, the law of conservation of atomic number and mass number is always obeyed and this helps us to calculate the number of `alpha and beta-` particles emitted in the reaction.
The number of `alpha - and beta-` particle emitted in nuclear reaction `._(90)Th^(288) rarr ._(83)Bi^(212)` are respectively

Options

A. 4, 1
B. 3, 7
C. 8, 1
D. 4, 7

Explanation

To solve the problem of determining the number of alpha and beta particles emitted during the nuclear reaction of thorium-288 to bismuth-212, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Elements**:
- The initial element is Thorium (Th) with an atomic number of 90 and mass number of 288: \(_{90}^{288}Th\).
- The final element is Bismuth (Bi) with an atomic number of 83 and mass number of 212: \(_{83}^{212}Bi\).

2. **Calculate the Change in Mass Number**:
- The mass number decreases from 288 to 212.
- Change in mass number = \(288 - 212 = 76\).

3. **Calculate the Number of Alpha Particles Emitted**:
- Each alpha particle (\(\alpha\)) has a mass number of 4.
- To find the number of alpha particles emitted, divide the total change in mass number by the mass number of one alpha particle:
\[ \text{Number of alpha particles} = \frac{76}{4} = 19. \]

4. **Calculate the Change in Atomic Number**:
- Each alpha particle emitted decreases the atomic number by 2.
- Therefore, the total decrease in atomic number from the emission of 19 alpha particles is:
\[ \text{Decrease in atomic number} = 19 \times 2 = 38. \]
- The new atomic number after emission of alpha particles:
\[ 90 - 38 = 52. \]

5. **Determine the Number of Beta Particles Emitted**:
- The final atomic number of Bismuth is 83.
- The difference in atomic number after alpha emissions and before reaching Bismuth:
\[ \text{Difference} = 83 - 52 = 31. \]
- Each beta particle (\(\beta\)) increases the atomic number by 1.
- Therefore, the number of beta particles emitted is:
\[ \text{Number of beta particles} = 31. \]

6. **Final Answer**:
- The number of alpha particles emitted is 19, and the number of beta particles emitted is 31.

### Summary of the Solution:
- Number of alpha particles emitted: **19**
- Number of beta particles emitted: **31**