JAMB - Chemistry (2017 - No. 21)
\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + alpha particle
226
220
227
222
Giải thích
\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + \(^4_{2}He\)
\(^4_{2}He\) = alpha particle
considering the summation of the mass number
226 = x + 4
x = 226 - 4
x = 222
\(^4_{2}He\) = alpha particle
considering the summation of the mass number
226 = x + 4
x = 226 - 4
x = 222