Question

11) (a) Provide your best estimate of the number of "C atoms in a typical 70 kg human. Show enough of your work to know how your arrived at your answer. (4 pts) Number of atoms in a typical 70 kg human = (b) Estimate how many "C decays occur per day in a typical 70 kg human. (4 pts) Show enough of your work to know how your arrived at your answer. Number of decays occur per day in a typical 70 kg human = (e) Explain how "c nuclei that undergo B-minus decay produce x-rays. (3 pts)

Answer 1

11)a

Person weight = 70Kg

The carbon content in a body is 23% ; 23% * 70 = 16.1 Kg or 16100 grams of carbon.

16100 X 1 mole per 23 grams = 700 moles of Carbon.

Or 700 x 6.022 x 1023 = 4.215 x 1026 atoms of carbon in the person (based on the carbon atoms against total atoms in a body)

and there is only one C14 per 1012 C12 so the total C14 in the person would be having 4.215 X 1014 C14 atoms.

Number of C14 in a typical 70Kg human would be having 4.215 X 1014 C14 atoms.

b) the number of C14 in a day in atypical 70Kg human would be as follows

Rate of decay = rate of decay X number of radioactive atoms

3.8359 x 10-12 per second X 4.215x1014 atoms = 1617 atoms per second will decay, and for a day it is as follows

No:of seconds in a day = 86400

Therefore the decay per day = 1617 x 86400 = 139,708,800 C14 atoms per day.

c) C14 nuclei that undergo $\beta$ - decay produce X-rays:

In beta minus $\beta$ - decay, C14 atoms decay into nitrogen - N14 atoms wherein, neutron is converted to a proton, which gives an electron and an electron anti-neutrino resulting in emitting electromagnetic radiation or X rays in this case.