JAMB - Physics (1994 - No. 37)
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
Selitys
\(^{235}_{92}U\) \(\to\) 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
\(^{b}_{a}Y\) = \(^{219}_{87}Y\)
a + (4 + 2) + (3x - 1) = 92 \(\Rightarrow\) a = 87
b + (4 x 4) + (3 x 0) = 235 \(\Rightarrow\) b = 219
\(^{b}_{a}Y\) = \(^{219}_{87}Y\)