"The half-life of Au198 is 2.7 days. The activity of 1.50 mg of Au198 if its atomic weight is 198 g mol$-$1 is, (Na = 6 $\times$ 1023/mol)."

Question

Accepted Answer

"Activity, $A = \lambda N$

Where, $N = n{N_A}$

${N_A} = 6 \times {10^{23}}$/mol

$= {{6 \times {{10}^{23}}} \over {3.7 \times {{10}^{10}}}}$ Ci

Here, ${A_0} = \lambda {N_0}$

We know,

${T_{1/2}} = {{\ln (2)} \over \lambda }$

$\Rightarrow \lambda = {{\ln (2)} \over {{T_{1/2}}}}$

$= {{\ln (2)} \over {2.7 \times 3600 \times 24}}$

$\therefore$ ${A_0} = {{\ln (2)} \over {2.7 \times 3600 \times 24}} \times$ $n \times {N_A}$

$$ = {{\ln (2)} \over {2.7 \times 3600 \times 24}} \times {{1.5 \times {{10}^{ - 3}}} \over {198}} \times {{6 \times {{10}^{23}}} \over {3.7 \times {{10}^{10}}}}$$

$= 366$ Ci

$\therefore$ Nearest answer is 357 Ci."

The half-life of Au¹⁹⁸ is 2.7 days. The activity of 1.50 mg of Au¹⁹⁸ if its atomic weight is 198 g mol^$-$1 is, (N_a = 6 $\times$ 10²³/mol).

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The half-life of Au198 is 2.7 days. The activity of 1.50 mg of Au198 if its atomic weight is 198 g mol$-$1 is, (Na = 6 $\times$ 1023/mol).

Solution

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The half-life of Au¹⁹⁸ is 2.7 days. The activity of 1.50 mg of Au¹⁹⁸ if its atomic weight is 198 g mol^$-$1 is, (N_a = 6 $\times$ 10²³/mol).