An atom 23892X decays by alpha particle emission to give an atom Y. The atomic number and mass number of Y are
-
A.
90 and 234 respectively -
B.
91 and 238 respectively -
C.
92 and 236 respectively -
D.
93 and 238 respectively
Correct Answer: Option A
Explanation
23892 X 234 90Y + 4 2 He Atomic number is 90 while mass number is 234