WAEC - Chemistry (2006 - No. 3)
An atom 23892X decays by alpha particle emission to give an atom Y. The atomic number and mass number of Y are
90 and 234 respectively
91 and 238 respectively
92 and 236 respectively
93 and 238 respectively
स्पष्टीकरण
23892 X 234 90Y + 4 2 He Atomic number is 90 while mass number is 234