Carbon-14 has a half-life of approximately 5,730 years. This exponential decay can be modeled with the function N(t) = N0. If an organism had 200 atoms of carbon-14 at death, how many atoms will be present after 14,325 years? Round the answer to the nearest hundredth.

This is of the form f=ir^t. We are given that the half life is 5730 years so we can say:

1=2r^5730

1/2=r^5730 taking natural log of both sides

ln0.5=5730lnr

lnr=(ln0.5)/5730 raising e to the power of both sides

r=e^((ln0.5)/5730)

r≈0.999879

so we have:

f=200(0.999879)^14325

f≈35.34

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tutorareej tutorareej · Answer 2 · 2021-11-10T14:06:36+01:00

After the period of 14,325 years, the amount of Carbon-14 remained is 35.356 atoms.

The half-life of the substance can be defined a the time required for the substance to reduce to half of the concentration.

The half-life can be calculated as:

[tex]\rm N_t\;=\;N_0\;(\dfrac{1}{2} )^\dfrac{t}{t_1_/_2}[/tex] ........ (i)

where [tex]\rm N_t[/tex] is the quantity of substance remaining

[tex]\rm N_0[/tex] is the initial quantity

t is the time of decay

[tex]\rm t_\dfrac{1}{2}[/tex] is the half-life of the substance.

In the question, [tex]\rm N_0[/tex] = 200, t = 14,325 years, and [tex]\rm t_\dfrac{1}{2}[/tex] = 5730 years.

Substituting the values in equation (i),

[tex]\rm N_t\;=\;200\;(\dfrac{1}{2} )^\dfrac{14,325}{5,730}[/tex]

[tex]\rm N_t\;=\;200\;(\dfrac{1}{2} )^2^.^5[/tex]

[tex]\rm N_t\;=\;200\;\times\;0.17678[/tex]

[tex]\rm N_t[/tex] = 35.356 atoms.

After the period of 14,325 years, the amount of Carbon-14 remained is 35.356 atoms.

For more information, refer to the link:

https://brainly.com/question/24710827