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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
\(^{235}_{92}U\) \(\to\) 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
\(^{b}_{a}Y\) = \(^{219}_{87}Y\)
Source: Myschool Past Questions Archive (Record #27936)