Answer:
A = 0.325 Bq
Explanation:
given,
half life of ¹⁴C = 5730 years
fixed fraction 1.30 × 10⁻¹² of ¹²C
half life = 5730 years
T_{1/2} = 5730 x 365 x 24 x 60 x 60
= 1.807 x 10¹¹ s
radioactive decay constant
λ = [tex]\dfrac{0.693}{T_{1/2}}[/tex]
λ = [tex]\dfrac{0.693}{1.807\times 10^{11}}[/tex]
λ = 3.835 x 10⁻¹² /s
number of atom
mass m = 1.30 g
[tex]n = \dfrac{m_0}{M}\times 6.022 \times 10^{23}[/tex]
[tex]n = \dfrac{1.30}{12}\times 6.022 \times 10^{23}[/tex]
n = 6.524 x 10²²
Number of ¹⁴C atoms in 1.3 g of sample N ' =
= 1.30 × 10⁻¹² x 6.524 x 10²²
= 8.482 x 10¹⁰
Required activity, A = λ N
A = 3.835 x 10⁻¹² x 8.482 x 10¹⁰
A = 0.325 Bq
