Question 250: Atoms & Nuclei

Question

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 $$	imes$$ 1023/mol).

Accepted Answer

Activity, $$A = \lambda N$$
Where, $$N = n{N_A}$$
$${N_A} = 6 	imes {10^{23}}$$/mol
$$ = {{6 	imes {{10}^{23}}} \over {3.7 	imes {{10}^{10}}}}$$ Ci
Here, $${A_0} = \lambda {N_0}$$
We know,
$${T_{1/2}} = {{\ln (2)} \over \lambda }$$
$$ \Rightarrow \lambda  = {{\ln (2)} \over {{T_{1/2}}}}$$
$$ = {{\ln (2)} \over {2.7 	imes 3600 	imes 24}}$$
$$	herefore$$ $${A_0} = {{\ln (2)} \over {2.7 	imes 3600 	imes 24}} 	imes $$ $$n 	imes {N_A}$$
$$ = {{\ln (2)} \over {2.7 	imes 3600 	imes 24}} 	imes {{1.5 	imes {{10}^{ - 3}}} \over {198}} 	imes {{6 	imes {{10}^{23}}} \over {3.7 	imes {{10}^{10}}}}$$
$$ = 366$$ Ci
$$ 	herefore $$ Nearest answer is 357 Ci.

Answer

240 Ci

Answer

357 Ci

Answer

252 Ci

Answer

535 Ci

Atoms & Nuclei

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).

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