A radioactive substance decays according to the formula A=A0e^kt
where
A0 is the initial amount of substance (in grams)
A is the amount of substance(in grams) after t years
k is a constant
The half-life of the substance is 10 years.If we begin with 20g of the substance,how much will be left after 5 years?
you can go the e^kt route, but it's more transparent to use base 2, since we're talking about half-lives. 2^-1 = 1/2, so
A = Ao*2^(-t/10)
after 5 years, you have
A = 20*2^(-5/10) = 14.14g
Of course, since 2 = e^ln2, that means you could use
(e^ln2)^(-t/10) = e^(-t*ln2/10) = e^(-0.0693t)
how much of the substance would be present after 5 years
e=1/2 since we are talking about half-lives
AO=20g
constant,k=elapsed time(t)/half-life
=5 years/10 years
=1/2
IF A=AOe^1/2
=20g(1/2)^1/2
=14.14 g
thats the solution happy dance
Thank you so much you helped
To find out how much of the substance will be left after 5 years, we can use the given formula A = A₀e^(kt).
Given:
Initial amount of substance, A₀ = 20g
Time, t = 5 years
Half-life, t₁/₂ = 10 years
First, let's find the value of the constant k using the half-life formula.
The half-life formula is:
t₁/₂ = (ln 2) / k
Substituting the given half-life, we get:
10 = (ln 2) / k
To solve for k, we can rearrange the equation as:
k = (ln 2) / 10
Now that we have the value of k, we can substitute it and the given values into the formula A = A₀e^(kt).
A = 20 * e^((ln 2/10) * 5)
Let's calculate the expression inside the exponential first, (ln 2/10) * 5:
(ln 2/10) * 5 ≈ 0.115
Then, we can continue with the equation:
A = 20 * e^0.115
Using a calculator or computer, we can evaluate e^0.115 to be approximately 1.122.
Therefore,
A ≈ 20 * 1.122 ≈ 22.44
After 5 years, approximately 22.44g of the substance will be left.