A lab technician had 58 grams of a radioactive substance. Ten days later, only 52 grams remained. Find the half-life of the radioactive substance.
if the half-life is n days, then the amount remaining after t days can be modeled by
58(1/2)^(t/n)
So, to find n, we just have to solve
58(1/2)^(10/n) = 52
(1/2)^(10/n) = 52/58 = 0.8966
10/n = log(52/58)/log(1/2)
10/n = 0.1575
n = 63.5
So, the half-life is about 63.5 days
To find the half-life of a radioactive substance, we can use the formula:
\(\text{{Half-life}} = \frac{{\ln(2)}}{{k}}\),
where \(k\) is the decay constant.
In this case, we are given the initial quantity of the substance (58 grams) and the quantity remaining after a certain number of days (52 grams). We can use this information to find the decay constant.
First, let's calculate the decay constant using the formula:
\(N(t) = N_0 \cdot e^{-kt}\),
where \(N(t)\) is the quantity remaining after time \(t\), \(N_0\) is the initial quantity, and \(e\) is the base of the natural logarithm.
Substituting the given values, we get:
\(52 = 58 \cdot e^{-10k}\).
Next, let's solve this equation for \(k\). Divide both sides by 58:
\(\frac{{52}}{{58}} = e^{-10k}\).
Take the natural logarithm of both sides:
\(\ln\left(\frac{{52}}{{58}}\right) = -10k\).
Evaluate the natural logarithm:
\(\ln\left(\frac{{52}}{{58}}\right) \approx -0.1056 = -10k\).
Now solve for \(k\) by dividing both sides by -10:
\(k \approx \frac{{-0.1056}}{{-10}} = 0.01056\).
Finally, substitute this value of \(k\) into the half-life formula:
\(\text{{Half-life}} = \frac{{\ln(2)}}{{0.01056}}\).
Evaluate this expression to find the half-life of the radioactive substance.