Radium 226, which decays according to the function A(t)=A0e^kt where t is the time in years, has a half life of 1612 years. Find k.
1/2 = e^(kt)
ln(1/2) = 1612 k
To find the value of k in the exponential decay equation A(t) = A₀e^(kt), we can use the half-life information given.
The half-life of a substance is the time it takes for half of the initial quantity to decay. In this case, the half-life of Radium-226 is 1612 years.
In the exponential decay equation, when t = half-life (t₁/₂), A(t) becomes half of the initial quantity A₀. So we have:
A(t₁/₂) = A₀/2
Substituting the values into the equation, we get:
A₀e^(k(t₁/₂)) = A₀/2
Canceling out A₀, we have:
e^(k(t₁/₂)) = 1/2
To solve for k, we need to isolate it. Taking the natural logarithm (ln) of both sides of the equation gives:
ln(e^(k(t₁/₂))) = ln(1/2)
Using the property of logarithms that ln(e^x) = x, we simplify further:
k(t₁/₂) = ln(1/2)
Now, we can divide both sides of the equation by (t₁/₂) to isolate k:
k = ln(1/2) / (t₁/₂)
Substituting the given value of t₁/₂ = 1612 years, we can calculate the value of k using a calculator or computer program:
k ≈ ln(1/2) / 1612
Hence, to find the value of k for Radium-226, calculate ln(1/2) and divide it by 1612.