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Question

In a radioactive decay chain, ​23290Th nucleus decays to 21282Pb nucleus. Let Nα ​ and Nβ ​ be the number of α and β particles respectively, emitted in this decay process. Which of the following statements is (are) true?

A
Nα=5
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B
Nα=6
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C
Nβ=2
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D
Nβ=4
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Solution

The correct option is C Nβ=2
We are given that
23290Th 21282Pb+42He+β
23290Th 21282Pb+42He+1e
Let us conserve the mass on both the sides
232=212+x×4
x=5
Thus, the number of alpha particles = 5
In order to find the number of beta particles, we also need to conserve the charge on both sides
90=82+5×2+y(1)
y=2
No. of beta particles = 2

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