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

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