Radio active decay: Half life

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, $t_{0.5}$.
$$\begin{eqnarray*} & & \frac{N_0}{2}=N_0e^{-\lambda t_{0.5}}  \\ \rightarrow & & \frac{1}{2} = e^{-\lambda t_{0.5}}  \\ \rightarrow & & 2 = e^{\lambda t_{0.5}}  \\ \rightarrow & & \lambda t_{0.5} = \ln 2  \\ \rightarrow & & t_{0.5} = \frac{\ln 2}{\lambda} \end{eqnarray*}$$
Thus, the half life is given by $\frac{0.693}{\lambda}$.

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