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

Correct answer is A

$^{235}_{92}U$ $\to$ 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

$^{b}_{a}Y$ = $^{219}_{87}Y$

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