A certain radioisotope of \(^{235}_{92}U\) emits four alpha particles and three beta particles. The mass number and the atomic number of the resulting elements respectively are
-
219 and 87
- 84 and 223
- 223 and 87
- 219 and 81
The correct answer is: A
Explanation
\(^{235}_{92}U\) \(\to\) 4\((^{4}_{2}\alpha)\) + 3\((^{0}_{-1}e)\) + \(^{b}_{a}y\)a + (4 + 2) + (3x - 1) = 92 \(\Rightarrow\) a = 87
b + (4 x 4) + (3 x 0) = 235 \(\Rightarrow\) b = 219
\(^{b}_{a}Y\) = \(^{219}_{87}Y\)