Question

Half-Life The radioactive decay of \(\mathrm{Sm}-151\) (an isotope of samarium) can be modeled by the differential equation \(d y / d t=\) \(-0.0077 y,\) where \(t\) is measured in years. Find the half-life of \(\mathrm{Sm}-151\).

Short answer

After calculating the expression for \(t\), we find that the half-life of the isotope Sm-151 is approximately 90 years.

Step-by-step solution

Identifying the differential equation and initial amount

The differential equation that is given is \( \frac{dy}{dt} = -0.0077y \). The equation has the initial amount of the isotope, \(y(0) = y_0\).

Solution of differential equation

Solving the differential equation we get \(y(t) = y_0e^{-0.0077t}\) where \(e\) is a mathematical constant approximately equal to 2.71828, \(t\) is the time in years and \(y_0\) is the initial amount of the isotope.

Applying half-life definition

The half-life is the time it takes for half of the isotope to decay, i.e, \(y(t) = 0.5y_0\). Substituting this into our solution, we get \(0.5y_0 = y_0e^{-0.0077t}\).

Solving for t

Simplifying the equation we get \(0.5 = e^{-0.0077t}\). Taking the natural logarithm of both sides, we get \(-0.0077t = ln(0.5)\). Solving for \(t\) gives us \(t = \frac{ln(0.5)}{-0.0077}\).