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In a slow reaction, heat is being evolved at a rate about 10mW in a liquid. If the heat were being generated by the decay of ${}^{32}P$ , a radioactive isotope of phosphorus that has half-life of 14 days and emits only beta-particles with a mean energy of 700KeV, estimate the number of ${}^{32}P$ atoms in the liquid. Express your answer in form of $A \times 10^{17}$ and fill 5A in OMR sheet. [Take : l n 2 = 0.675]

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answer is 8.

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Detailed Solution

$P = 700 \times 1 0^{3} \times 1.6 \times 1 0^{- 19} \times \frac{d N}{d t} = 10 \times 1 0^{- 3};$

$\frac{d N}{d t} = \frac{10^{- 2}}{10^{- 14}} \times \frac{1}{7 \times 16} = \frac{10^{12}}{11.2} = λ N_{0}$

$∴ λ = \frac{\ln 2}{14 \times 86400} \Rightarrow N_{0} = \frac{14 \times 86400 \times 10^{12}}{11.2 \ln 2} = 160 \times 1 0^{15}$

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