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 \( k \) ?

To find the half-life of Sm-151 from the given differential equation \( \frac{dy}{dt} = -0.0077y \), we can use the relationship for half-life, which is \( t_{1/2} = \frac{\ln(2)}{k} \). Here, \( k = 0.0077 \). Therefore, the half-life is calculated as \( t_{1/2} = \frac{\ln(2)}{0.0077} \approx 90.0 \) years. That means in about 90 years, half of the Sm-151 will have decayed away! For Np-240, knowing that the half-life is 65 minutes allows us to find the decay constant \( k \) in a similar manner. Using the same half-life formula, \( t_{1/2} = \frac{\ln(2)}{k} \), we can rearrange this to find \( k \) where \( k = \frac{\ln(2)}{t_{1/2}} \). Substituting in 65 minutes gives us \( k \approx \frac{0.693}{65} \approx 0.01065 \) per minute. This means Np-240 has a quick decay for sure!