One mole of radium has an activity of $\frac{1}{3.7}$ kilo curie. Its decay constant is (Avagadro number $=6…

One mole of radium has an activity of $\frac{1}{3.7}$ kilo curie. Its decay constant is (Avagadro number $=6 \times 10^{23} \mathrm{~mol}^{-1}$ )
  1. $\frac{1}{6} \times 10^{-10} s^{-1}$
  2. $10^{-10} s^{-1}$
  3. $10^{-11} s^{-1}$
  4. $10^{-8} s^{-1}$

Solution

For radium, $\mathrm{n}=1$ mole, $\mathrm{A}=\frac{1}{3.7} \mathrm{k}$ curie $\therefore \quad \mathrm{n}=\frac{\mathrm{N}}{\mathrm{~N}_{\mathrm{A}}} \Rightarrow \mathrm{~N}=\mathrm{nN}_{\mathrm{A}}=6 \times 10^{23}$ $\therefore$ Decay constant, $\lambda=\frac{\mathrm{A}}{\mathrm{N}}=\frac{\frac{1}{3.7} \times 10^3 \times 3.7 \times 10^{10}}{6 \times 10^{23}}$ $=\frac{1}{6} \times 10^{-10} \mathrm{~s}^{-1}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

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