Question

Carbon-14 is a radioactive isotope that has a half-life of 5,730 years. Approximately how many years will it take for carbon-14 to decay to 10 percent of its original amount?

Answer 1

Answer:

Carbon-14 will take 19,035 years to decay to 10 percent.

Step-by-step explanation:

Exponential Decay Function

A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay.

An exponential decay can be described by the following formula:

$N(t)=N_{0}e^{-\lambda t}$

Where:

No  = The quantity of the substance that will decay.

N(t) = The quantity that still remains and has not yet decayed after a time t

$\lambda$     = The decay constant.

One important parameter related to radioactive decay is the half-life:

$\displaystyle t_{1/2}=\frac {\ln(2)}{\lambda }$

If we know the value of the half-life, we can calculate the decay constant:

$\displaystyle \lambda=\frac {\ln(2)}{ t_{1/2}}$

Carbon-14 has a half-life of 5,730 years, thus:

$\displaystyle \lambda=\frac {\ln(2)}{ 5,730}$

$\lambda=0.00012097$

The equation of the remaining quantity of Carbon-14 is:

$N(t)=N_{0}e^{-0.00012097\cdot t}$

We need to calculate the time required for the original amout to reach 10%, thus N(t)=0.10No

$0.10N_o=N_{0}e^{-0.00012097\cdot t}$

Simplifying:

$0.10=e^{-0.00012097\cdot t}$

Taking logarithms:

$ln 0.10=-0.00012097\cdot t$

Solving for t:

$\displaystyle t=\frac{log 0.10}{-0.00012097}$

$t\approx 19,035\ years$

Carbon-14 will take 19,035 years to decay to 10 percent.

Answer 2

Answer:

In photo below

Explanation:

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