We will find that the half-life of the given radioactive element is 5.43 years.
Let's see how to solve this:
We define the half-time of a quantity A, which decays exponentially, as the time such that the initial quantity reduces to its half.
Here we know that, after 4 years, 40% of a radioactive element decays.
So if the decay equation is:
D(t) = A*e^{k*t}
Where A is the initial amount, k is a constant and t is the time in years.
We know that after 4 years we will have:
D(4) = 0.6*A = A*e^{k*4}
Now we can solve this for k:
0.6*A = A*e^{k*4}
0.6 = e^{k*4}
Now we can apply the natural logarithm to both sides to get:
ln(0.6) = ln(e^{k*4}) = k*4
ln(0.6)/k = 4 = -0.1277
Then the decay equation is:
D(t) = A*e^{-0.1277*t}
To find the half-time, we must solve:
D(t) = 0.5*A = A*e^{-0.1277*t}
0.5 = e^{-0.1277*t}
ln(0.5) = ln(e^{-0.1277*t}) = -0.1277*t
ln(0.5)/(-0.1277) = t = 5.43
So, concluding, the half-life of the radioactive element is 5.43 years.
If you want to learn more, you can read:
https://brainly.com/question/16387602
