Answer:
Explanation:
From the information given:
The half-life [tex]t_{1/2}[/tex] = 1.6103 years
The no. of the initial nuclei [tex]N_o[/tex] = [tex]4.00 \times 10^6[/tex]
Using the formula:
[tex]N = N_o exp(-\lambda t)[/tex]
where;
decay constant [tex]\lambda = \dfrac{In2}{1.6*10^3} y^{-1}[/tex]
∴
[tex]N = N_o exp ( \dfrac{-In2}{1.6*10^3}\times 4.4 \times 10^3)[/tex]
[tex]N = N_o exp (- 1.906154747)[/tex]
The number of radium nuclei N = 5.94 × 10¹⁵
The initial activity[tex]A_o = \lambda N_o[/tex]
[tex]A_o =(\dfrac{In (2)}{1.61\times 10^3 \times 365 \times 24 \times 3600}\times 4.00 \times 10^{16})[/tex]
[tex]A_o =546075.8487 \ Bq[/tex]
Since;
1 curie = 3.7 × 10¹⁰ Bq
Then;
[tex]A_o =\dfrac{546075.8487 }{3.7\times 10^{10}}[/tex]
[tex]A_o = 1.47588 \times 10^{-5}Ci[/tex]
[tex]A_o = 14.7588 \ \mu Ci[/tex]
c) The activity at a later time is:
[tex]=5.94 \times 10^{15}( \dfrac{In (2)}{1.60 \times 10^3 \times 365\times 24 \times 3600})[/tex]
[tex]= 81599.09018 \ Bq \\ \\ = \dfrac{81599.09018}{3.7\times 10^{10}} \ Ci \\ \\ = 2.20538 \times 10^6 \ Ci \\ \\ = 2.20538 \ \mu Ci[/tex]
