An element X of an atomic number 88 and mass number 226 decays to form an element Z by emitting two beta particles and an alpha particle. Z is represented by
-
A.
(^{222}_{82}Z) -
B.
(^{222}_{88}Z) -
C.
(^{226}_{86}Z) -
D.
(^{226}_{80}Z)
Correct Answer: Option B
Explanation
(^{226}_{88}X) (to) ((^{0}_{-1}e)) + (^{4}_{2}alpha) + (^{b}_{a}X)
a + 2 + (-2) = 88 a (Rightarrow) a = 88
b + 4 + 0 = 266 b (Rightarrow) b = 222