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
A. 219 and 87 B. 84 and 223 C. 223 and 87 D. 219 and 81
Correct Answer: 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 \(^{b}_{a}Y\) = \(^{219}_{87}Y\)
