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: Option 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)