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

View answer & explanation

Correct answer is A

23892 X 234 90Y + 4 2 He Atomic number is 90 while mass number is 234

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