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(^{226}_{88}Ra) → (^x_{86}Rn) + alpha particle

(^{226}_{88}Ra) → (^x_{86}Rn) + alpha particle

  • A.
    226
  • B.
    220
  • C.
    227
  • D.
    222
Correct Answer: Option D
Explanation

(^{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

There is an explanation video available below.

Explanation Video