226 88 Ra x 86 Rn alpha particle - JAMB Chemistry 2017 Question
(^{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
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