A radioactive isotope, selenium, used in the creation of medical images of the pancreas,
has a half-life of 119.77 days. If 100 milligrams are given to a patient, how many milligrams
are left after 20 days?
7.
The Lualailua Hills Quadrangle of the East Maui (Haleakala) volcano on the island of
Maui in Hawaii is no longer active. To find out the date of the last eruption, scientists
conducted a chemical analysis of samples from the volcano area. The samples
contained approximately 62.31% of its original carbon-14. How long ago was the last
eruption of the volcano? (Use 5730 years for the half-life of carbon-14.)

6) The amount of substance left after 20 minutes is; A = 89.07 mg

7) The last time of the volcanic eruption is; 3910.11 years.

6) The formula for radioactive decay is;

A = A_o*2^(-t/h)

Where;

A = Amount remaining after time t

A_o = Initial amount

t = time of decay

h = half-life of substance

We are given;

A_o = 100 mg

h = 119.77 days

t = 20 days

Thus;

A = 100 × 2^(-20/119.77)

A = 89.07 mg

7) We are given;

A/A_o = 62.31% = 0.6231

h = 5730 years

Thus;

A/A_o = 0.6231 = 2^(-t/5730)

Thus;

Log 0.6231 = -(t/5730) log 2

t = (−0.2054/-0.301) × 5730

t ≈ 3910.11 years

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First problem: 89.07 grams

Second problem: 3910.15 days

Step-by-step explanation:

We can solve these problems with the exponencial equation:

P = Po * (1+r)^t

Where P is the final value, Po is the inicial value, r is the rate and t is the time.

In the first problem, the half-life is 119.77 days and Po = 100, so we have that:

P = 100 * (1-0.5)^(t/119.77)

So after 20 days, we will have:

P = 100 * (1-0.5)^(20/119.77) = 89.07 grams

For the second problem, we have the half-life of 5730 days, and the final value over the inicial value is 62.31%, so:

P/Po = (1-0.5)^(t/5730)

0.6231 = 0.5^(t/5730)

log(0.6231) = log(0.5^(t/5730))

-0.4730 = (t/5730)*log(0.5)

-0.4730 = (t/5730)*(-0.6931)

(t/5730) = 0.6824

t = 0.6824 * 5730 = 3910.15 days