Radon undergoes decay by α-emission 88Rn222⟶t1/2=3.8 decay 84Po218+2He4Which of the following statements will be true of this decay process?

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If the initial amount of radon was I mg, the amount of radon left after 11.4 days will be 0. 125 mg.

Activity of radon after 7.6 days will be N0×5.3×10−7s−1 when N0 is the original number of atoms of the radon.

The decay constant of radon is 2.1 × 10-6 s-1

60% of the radon will decay in 5 days approximately

answer is A.

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Detailed Solution

N=N012n=1123= 0.125 mg.n = Number of half-life = 11.4/3.8 = 3Activity after 7.6 days = K × Remaining number of atoms =0.6933.8×24×3600×N04=5.3×10−7dissec−1K=0.6933.8×24×3600=2.1×10−6sec−1K=2.303tlog⁡N0N0.6933.8=2.3035log⁡100NN = 40∴% decay = 100 - 40 = 60

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