Question

The mass percent of carbon in a typical human is 18%, and the mass percent of 14C in natural carbon is 1.6 × 10-10%. Assuming a 190.-lb person, how many decay events per second occur in this person due exclusively to the β-particle decay of 14C? For 14C, t1/2 = 5730 years

Answer 1

Answer:

4066 decay/s

Explanation:

Given that:-

The weight of the person is:- 190 lb

Also, 1 lb = 453.592 g

So, weight of the person = 86182.6 g

Also, given that carbon is 18% in the human body. So,

$\%\ Carbon=\frac{18}{100}\times 86182.6\ g=15512.868\ g$

Carbon-14 is $1.6\times 10^{-10}\ \%$ of the carbon in the body. So,

$\%\ Carbon-14=\frac{1.6\times 10^{-10}}{100}\times 15512.868\ g=2.48\times 10^{-8}\ g$

Also,

14 g of Carbon-14 contains  $6.023\times 10^{23}$ atoms of carbon-14

So,  

$2.48\times 10^{-8}\ g$ of Carbon-14 contains  $\frac{6.023\times 10^{23}}{14}\times 2.48\times 10^{-8}$ atoms of carbon-14

Atoms of carbon-14 =  $1.07\times 10^{15}$

Given that:

Half life = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,  

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac{ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.00012 years⁻¹

Also, 1 year = $3.154\times 10^7$ s

So, The rate constant, k = $\frac{0.00012}{3.154\times 10^7}$ s⁻¹ = $3.8\times 10^{-12}\ s^{-1}$

Thus, decay events per second = $K\times atoms decayed$ = $3.8\times 10^{-12}\times 1.07\times 10^{15}\ decay/s$ = 4066 decay/s