21. Half-Life The radioactive decay of 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 Sm-151. 22. Half-Life An isotope of neptunium (Np-240) has a half-life of 65 minutes. If the decay of \( \mathrm{Np}-240 \) is modeled by the differential equation \( d y / d t=-k y \), where \( t \) is measured in minutes, what is the decay constant \( k \) ?

1. The differential equation is \[ \frac{dy}{dt} = -0.0077\, y. \] Its solution can be written as \[ y(t) = y_0 e^{-0.0077t}, \] where \( y_0 \) is the initial quantity. The half-life \( T_{1/2} \) is defined by \[ y(T_{1/2}) = \frac{y_0}{2}. \] Substituting into the solution: \[ y_0 e^{-0.0077 T_{1/2}} = \frac{y_0}{2}. \] Dividing both sides by \( y_0 \), \[ e^{-0.0077 T_{1/2}} = \frac{1}{2}. \] Taking the natural logarithm on both sides, \[ -0.0077 T_{1/2} = \ln\left(\frac{1}{2}\right) = -\ln 2. \] Solve for \( T_{1/2} \): \[ T_{1/2} = \frac{\ln 2}{0.0077}. \] Numerically, using \(\ln 2 \approx 0.6931\): \[ T_{1/2} \approx \frac{0.6931}{0.0077} \approx 90.0 \text{ years}. \] 2. The differential equation for the decay of \( \mathrm{Np}-240 \) is given by: \[ \frac{dy}{dt} = -k\, y. \] Its solution is \[ y(t) = y_0 e^{-k t}. \] Given that the half-life is 65 minutes, we have: \[ y(65) = \frac{y_0}{2}. \] So, \[ y_0 e^{-65k} = \frac{y_0}{2}. \] Cancel \( y_0 \) from both sides: \[ e^{-65k} = \frac{1}{2}. \] Taking the natural logarithm: \[ -65k = \ln \left( \frac{1}{2} \right) = -\ln 2. \] Thus, solving for \(k\): \[ k = \frac{\ln 2}{65}. \] Numerically, \(\ln 2 \approx 0.6931\): \[ k \approx \frac{0.6931}{65} \approx 0.01066 \text{ per minute}. \]