The amount of a radioactive substance remaining as it decays over time is A = A0(0.5)t/h ,where a represents the final amount
Ao represents the original amount, t represents the number of years, and h represents the half-life of the sibstance.
The half-life of the radioactive isotope carbon-14 is 5,730 years. Approximately how many years will it take a 50-gram mass of
carbon-14 to decay to 5 grams?

The amount of a radioactive substance remaining as it decays over time is A A005th where a represents the final amount Ao represents the original amount t repre class=

Respuesta :

The amount of a radioactive substance remaining as it decays over time is A = A0(0.5)t/h ,where a represents the final amount
Ao represents the original amount, t represents the number of years, and h represents the half-life of the sibstance.
The half-life of the radioactive isotope carbon-14 is 5,730 years. Approximately how many years will it take a 50-gram mass of
carbon-14 to decay to 5 grams?

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

What is the time of decay?

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:

[tex]N(t)=N_oe^{-\lambda t}[/tex]

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

[tex]\lambda[/tex]    = The decay constant.

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

[tex]t_{1/2}=\dfrac{Ln(2)}{\lambda}[/tex]

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

[tex]\lambda =\dfrac{ln(2)}{t_{1/2}}[/tex]

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

[tex]\lambda =\dfrac{ln(2)}{5730}[/tex]

[tex]\lambda[/tex] = 0.00012097

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

[tex]N(t)=N_oe^{-0.00012097t}[/tex]

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

[tex]0.10N_o=-N_oe^{-0.00012097t}[/tex]

Simplifying:

[tex]0.10=e^{0.00012097t}[/tex]

Taking logarithms:

ln(0.10) = -0.00012097t

Solving for t:

[tex]t=\dfrac{Log0.10}{-0.00012097}[/tex]

t = 19035 years

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

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