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

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:

 ........ (i)

where is the quantity of substance remaining

is the initial quantity

t is the time of decay

is the half-life of the substance.

In the question, = 200, t = 14,325 years, and = 5730 years.

Substituting the values in equation (i),

= 35.356 atoms.

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

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14 ft - 8 ft = 6 ft

Step-by-step explanation:

You basically just need to subtract it! :)