Let N β be the number of β particles emitted by 1 gram of Na24 radioactive nuclei (half life = 15 hrs) in 7.5 hours, N β is close to (Avogadro number = 6.023 × 1023 / g. mole): (a) 6.2 × 1021 (b) 7.5 × 1021 (c) 1.25 × 1022 (d) 1.75 × 1022

Answer: (b). Initial nuclei N₀ = 1/24 × 6.023 × 10²³ = 2.51 × 10²². 7.5 h is half of one half-life, so remaining fraction = 2-1/2 = 0.707. Each decay gives one β: Nβ = N₀(1 - 0.707) = 0.293 × 2.51 × 10²² ≈ 7.4 × 10²¹. So the answer is (b) 7.5 × 10²¹.

Let N β be the number of β particles emitted by 1 gram of Na24… | JEE Main 2015 (11 Apr, Shift 1)

Q92JEE Main 2015 · 11 APR · SHIFT 1Medium

Law of Radioactive Decay

Let N $\beta$ be the number of $\beta$ particles emitted by 1 gram of Na24 radioactive nuclei (half life = 15 hrs) in 7.5 hours, N $\beta$ is close to (Avogadro number = 6.023 $\times$ 1023 / g. mole):

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Answer: (b)

Initial nuclei $N_0 = \frac{1}{24} \times 6.023 \times 10^{23} = 2.51 \times 10^{22}$ .
$7.5$ h is half of one half-life, so remaining fraction $= 2^{-1/2} = 0.707$ .
Each decay gives one $\beta$ : $N_\beta = N_0(1 - 0.707) = 0.293 \times 2.51 \times 10^{22} \approx 7.4 \times 10^{21}$ .
So the answer is (b) $7.5 \times 10^{21}$ .

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Law of Radioactive Decay question from JEE Main 2015 (11 Apr, Shift 1)

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