Mathematics, 03.02.2020 09:54 ayeequeen6392

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.

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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

Answer from: silviamgarcia

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:

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

where $\rm N_t$ is the quantity of substance remaining

$\rm N_0$ is the initial quantity

t is the time of decay

$\rm t_\dfrac{1}{2}$ is the half-life of the substance.

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

Substituting the values in equation (i),

$\rm N_t\;=\;200\;(\dfrac{1}{2} )^\dfrac{14,325}{5,730}$

$\rm N_t\;=\;200\;(\dfrac{1}{2} )^2^.^5$

$\rm N_t\;=\;200\;\times\;0.17678$

$\rm N_t$ = 35.356 atoms.

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

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Answer from: Quest

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Answer from: Quest

The correct answer is -4

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