At time $t$, let say there are $N$ atoms of ${}^7\mathrm{Be}$ (radioactive). Then the net rate of formation of ${}^7\mathrm{Be}$ nuclei at this instant is,

                                                      $\frac{dN}{dt}=\frac{10^{-4}}{1\text{.}6\times 10^{-19}\times 1000}-\lambda N$

or                                                      $\frac{dN}{dt}=6\text{.}25\times 10^{11}-\lambda N$

or                                                      ${\int }_o^{N_0}\frac{dN}{6\text{.}25\times 10^{11}-\lambda N}={\int }_0^{3600}dt$

where $N_0$ are the number of nuclei at$t = 1 \mathrm{hour}$ or $3600 \mathrm{s}$.

∴                                                      $-\frac{1}{\lambda }\text{ln}\left(\frac{6\text{.}25\times 10^{11}-\lambda N_0}{6\text{.}25\times 10^{11}}\right)=3600$

$\lambda N_0$ = activity of ${}^7\mathrm{Be}$ at$t = 1 \mathrm{hour}$ $= 1.8 \times {10}^8$ disintegrations per second

∴                                                      $-\frac{1}{\lambda }\text{ln}\left(\frac{6\text{.}25\times 10^{11}-1\text{.}8\times 10^8}{6\text{.}25\times 10^{11}}\right)=3600$

∴                                                    $\lambda = 8.0 \times {10}^{-8} {\mathrm{s}}^{-1}$

Therefore, half-life                                    $t_{1/2}=\frac{0\text{.}693}{8\text{.}0\times 10^{-8}}=8\text{.}66\times 10^6 \mathrm{s}$

                                                                       $= 100.26$ days