Given,
[tex]\begin{gathered} N_0=75\text{ mg} \\ t=750\text{ hrs} \\ t_{\frac{1}{2}}=7.5\text{ days=7.5}\times24=180\text{ hrs} \end{gathered}[/tex]The reaction constant λ is given as,
[tex]\begin{gathered} \lambda=\frac{0.693}{t_{\frac{1}{2}}} \\ =\frac{0.693}{180\text{ hrs}} \\ =3.85\times10^{-3}hr^{-1} \end{gathered}[/tex]According to law of radioactive decay number of particle after time t is given as,
[tex]N=N_0e^{-\lambda t}[/tex]Substituting all known values,
[tex]\begin{gathered} N=75\text{ mg}\times e^{-(3.85\times10^{-3}\times750)} \\ =75\text{ mg}\times e^{-2.8875} \\ =75\text{ mg}\times0.0557 \\ =4.1775\text{ mg} \end{gathered}[/tex]Therefore, 4.1775 mg of the sample will be left after 750 hrs.
