Question

Solve this :

$$\begin{gathered} 70.\mathrm{A}\mathrm{sample}\mathrm{of}\mathrm{radioactive}\mathrm{element}\mathrm{containing}4\times {10}^{16}\mathrm{active}\mathrm{nuclei}.\mathrm{Half}\mathrm{life}\mathrm{of}\mathrm{element}\mathrm{is}10\mathrm{days},\mathrm{then}\mathrm{number}\mathrm{of}\mathrm{decayed}\mathrm{nuclei}\mathrm{after}30\mathrm{days}: \\ \left(1\right)0.5\times {10}^{16}\left(2\right)2\times {10}^{16} \\ \left(3\right)3.5\times {10}^{16}\left(4\right)1\times {10}^{16} \end{gathered}$$

Answer 1

$$\begin{gathered} Weknowthat \\ N=N_0{\left(\frac{1}{2}\right)}^n \\ wherenisnumberofhalflives \\ {N}_0=4\times {10}^{16} \\ Halflife=10days \\ 30days=3\times 10days=3halflives \\ N=4\times {10}^{16}\times {\left(\frac{1}{2}\right)}^3 \\ N=4\times \frac{1}{8}\times {10}^{16} \\ N=\frac{1}{2}\times {10}^{16}=0.5\times {10}^{16} \\ Numberofnucleidecayed \\ {N}_o-N=4\times {10}^{16}-0.5\times {10}^{16}=3.5\times {10}^{16} \end{gathered}$$