"The half-life of ${}^{198}Au$ is 3 days. If atomic weight of ${}^{198}Au$ is 198 g/mol then the activity of 2 mg of ${}^{198}Au$ is [in disintegration/second] :"

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Accepted Answer

"A = $\lambda$N

$$\lambda = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over {3 \times 24 \times 60 \times 60}}$$sec^$-$1 = 2.67 $\times$ 10^$-$6 sec^$-$1

N = Number of atoms in 2 mg Au

$= {{2 \times {{10}^{ - 3}}} \over {198}} \times 6 \times {10^{23}}$ = 6.06 $\times$ 10¹⁵

$A = \lambda N = 1.618 \times {10^{13}} = 16.18 \times {10^{12}}$ dps"

The half-life of ${}^{198}Au$ is 3 days. If atomic weight of ${}^{198}Au$ is 198 g/mol then the activity of 2 mg of ${}^{198}Au$ is [in disintegration/second] :

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The half-life of ${}^{198}Au$ is 3 days. If atomic weight of ${}^{198}Au$ is 198 g/mol then the activity of 2 mg of ${}^{198}Au$ is [in disintegration/second] :

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