Radon undergoes decay by $α$ -emission
$_{88} {Rn}^{222} {\overset{t_{1 / 2} = 3 .8 decay}{⟶}}_{84} {Po}^{218} +_{2} {He}^{4}$
Which 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 $N_{0} \times (5 .3 \times 10^{- 7}) s^{- 1}$ 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, B, C, D.

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Detailed Solution

$N = N_{0} {(\frac{1}{2})}^{n} = 1 {(\frac{1}{2})}^{3}$

= 0.125 mg.

n = Number of half-life = 11.4/3.8 = 3

Activity after 7.6 days = K × Remaining number of atoms $= \frac{0.693}{3.8 \times 24 \times 3600} \times \frac{N_{0}}{4} = 5.3 \times 10^{- 7} dis \sec^{- 1}$

$\begin{array}{l} K = \frac{0 .693}{3 .8 \times 24 \times 3600} = 2 .1 \times 10^{- 6} \sec^{- 1} \\ K = \frac{2 .303}{t} \log (\frac{N_{0}}{N}) \\ \frac{0 .693}{3 .8} = \frac{2 .303}{5} \log (\frac{100}{N}) \end{array}$