Question on: JAMB Chemistry - 2017
\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + alpha particle
A
226
B
220
C
227
D
222
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Correct Option: D
\(^{226}_{88}Ra\) → \(^x_{86}Rn\) + \(^4_{2}He\)
\(^4_{2}He\) = 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
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