Question: 

Find the half life for the process of radioactive decay. Assume that \(\lambda\) is the decay constant.

Answer:

The radioactive decay is understood as the comparison of the numbers of nucleus at time \(t\) and time \(t+\Delta t\). The difference of the numbers can be expressed as \(N(t)-N(t+\Delta t)\), and this is proportional to a rate, \(\lambda\).

\[  N(t)-N(t+\Delta t) = -\lambda N(t) \Delta t  \]

The right hand side indicates how the number decreases for \(\Delta t\). We can rearrange it as

\[  \frac{N(t)-N(t+\Delta t)}{\Delta t}=\frac{dN}{dt}=-\lambda N(t)  \]

This is a first order of differential equation with respect to time. Again, rearrange and integrate it as follows:

\[  \int^N_{N_0} \frac{dN}{N}=\int^{t}_{0} -\lambda dt  \]

This gives

\[  \ln \frac{N}{N_0} = -\lambda t \quad \rightarrow \quad N=N_0e^{-\lambda t}  \]

In order to find the half life, let \(N=\frac{N_0}{2}\). Then derive the time, \(t_{0.5}\).

\[ \frac{N_0}{2}=N_0e^{-\lambda t_{0.5}}  \]

\[  \rightarrow  \frac{1}{2} = e^{-\lambda t_{0.5}}  \]

\[  \rightarrow  2 = e^{\lambda t_{0.5}}  \]

\[  \rightarrow  \lambda t_{0.5} = \ln 2  \]

\[  \rightarrow  t_{0.5} = \frac{\ln 2}{\lambda}  \]

Thus, the half life is given by \(\frac{0.693}{\lambda}\).