JEE Advanced 2018 · Paper 2 · Physics Q5

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PhysicsMultiple correct+4 / -2Question 5 of 54

In a radioactive decay chain, ${}^{232}_{90}\mathrm{Th}$ nucleus decays to ${}^{212}_{82}\mathrm{Pb}$ nucleus. Let $N_\alpha$ and $N_\beta$ be the number of $\alpha$ and $\beta^-$ particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

N_\alpha = 5

N_\alpha = 6

N_\beta = 2

N_\beta = 4

Answer and worked solution

Official answer: A, C

α particles. Only α emission changes $A$ , by 4 each:
$232-212=20=4N_\alpha\implies N_\alpha=5.$

β⁻ particles. Five α lower $Z$ by 10. Each β⁻ raises it by 1:
$90-10+N_\beta=82\implies N_\beta=2.$

(A), (C)

Topics:Radioactive decay

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JEE Advanced 2018 · Paper 2, Physics question 5