(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

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(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 $T_{1/2} = 2.2 \times 10^9 \; s$,we find $\lambda = \frac{0.693}{2.2 \times 10^9} \approx 3.15 \times 10^{-10} \; s^{-1}$.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 $N = \frac{R}{\lambda} = \frac{10^{10}}{3.15 \times 10^{-10}} \approx 3.17 \times 10^{19}$ atoms.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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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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A
$3.17 \times 10^{20}$
B
$3.17 \times 10^{19}$
C
$3.17 \times 10^{17}$
D
$3.17 \times 10^{18}$

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Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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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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