If 40 milligrams of strontium-90 radioactively decays to 12 milligrams in 30 years, find its half-life (the number of years it takes until half of it remains). Use the formula A = p ⋅ e−kt, where p is the amount and A the (smaller) final amount.

Question

If 40 milligrams of strontium-90 radioactively decays to 12 milligrams in 30 years, find its half-life (the number of years it takes until half of it remains). Use the formula A = p ⋅ e−kt, where p is the amount and A the (smaller) final amount.

Answer 1

12 = 40 e^(-kt)

.3 = e^-30 k
ln .3 = -30 k
-1.20 = -30 k
k = .04013
so
.5 = e^-.04013 t
-.6931 = -.04013 t
t = 17.3 years

Answer 2

To find the half-life of strontium-90, we can use the formula A = p ⋅ e^(-kt), where A is the final amount, p is the initial amount, k is the decay constant, and t is the time.

In this case, we know that the initial amount (p) is 40 milligrams, and the final amount (A) is 12 milligrams after 30 years. Plugging these values into the equation, we get:

12 = 40 * e^(-kt)

To solve for k, we need to isolate it. Divide both sides of the equation by 40:

12/40 = e^(-kt)

0.3 = e^(-kt)

Now, take the natural logarithm (ln) of both sides:

ln(0.3) = ln(e^(-kt))

Using the property ln(e^x) = x, the equation simplifies to:

ln(0.3) = -kt

To find k, divide both sides of the equation by -t:

k = -ln(0.3) / t

Now that we have the value of k, we can use it to find the half-life. The half-life (t_1/2) is the time it takes for the amount to reduce to half its initial value. In this case, it is the time it takes for the amount to reduce from 40 milligrams to 20 milligrams.

Using the equation A = p ⋅ e^(-kt), we can set A = 20 milligrams, p = 40 milligrams, and solve for t:

20 = 40 * e^(-k * t_1/2)

Divide both sides of the equation by 40:

0.5 = e^(-k * t_1/2)

Take the natural logarithm (ln) of both sides:

ln(0.5) = ln(e^(-k * t_1/2))

Again, using the property ln(e^x) = x, the equation becomes:

ln(0.5) = -k * t_1/2

To find t_1/2, divide both sides of the equation by -k:

t_1/2 = -ln(0.5) / k

Now that we have the value of k, we can substitute it into the equation:

t_1/2 = -ln(0.5) / (-ln(0.3) / t)

Simplifying further:

t_1/2 = t * ln(0.5) / ln(0.3)

Calculating the values:

t_1/2 = 30 * ln(0.5) / ln(0.3) ≈ 45.816 years

Therefore, the half-life of strontium-90 is approximately 45.816 years.