An atom 23892X decays by alpha particle emission to give an atom Y. The atomic number and mass number of Y are

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

waec 2006 chemistry

The correct answer is A. 90 and 234 respectively

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

Previous question Next question