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One mole of radium has an activity of 1/3.7 kilo curie. Its decay constant will be

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a

1/6 x 10^–10 s^–1

b

10^–10 s^–1

c

10^–11 s^–1

d

10^–8 s^–1

answer is A.

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Detailed Solution

$\begin{array}{l} A = \frac{1}{3 .7} \times 10^{3} Ci \\ (∴ 1 Ci = 3 .7 \times 10^{10} dis / \sec) \\ A = \frac{1}{3 .7} \times 10^{3} \times 3 .7 \times 10^{10} = 10^{13} / \sec \\ From A = Nλ \Rightarrow λ = \frac{A}{N_{A}} = \frac{10^{13}}{6 \times 10^{23}} \\ ∴ λ = \frac{1}{6} \times 10^{- 10} s^{- 1} \end{array}$

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