The rate of radioactive disintegration at an | vedclass

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Question

(B) The relationship between half-life $T_{1/2}$ and decay constant $\lambda$ is given by $T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambd

Vedclass · Accepted Answer

$3.17 	imes 10^{19}$

Vedclass · Answer

(B) The relationship between half-life $T_{1/2}$ and decay constant $\lambda$ is given by $T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}$.Given $T_{1/2} = 2.2 	imes 10^9 \; s$,we find $\lambda = \frac{0.693}{2.2 	imes 10^9} \approx 3.15 	imes 10^{-10} \; s^{-1}$.The rate of disintegration $R$ is related to the number of radioactive atoms $N$ by the formula $R = \lambda N$.Given $R = 10^{10} \; s^{-1}$,we calculate $N = \frac{R}{\lambda} = \frac{10^{10}}{3.15 	imes 10^{-10}} \approx 3.17 	imes 10^{19}$ atoms.

The rate of radioactive disintegration at an instant for a radioactive sample of half-life $2.2 \times 10^9 \; s$ is $10^{10} \; s^{-1}$. The number of radioactive atoms in that sample at that instant is,

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