Question

The fraction of atoms of radioactive element that decays in 6 days is $\frac{7}{8}$. The fraction that decays in 10 days will be

• A. $\frac{77}{80}$
• B. $\frac{71}{80}$
• C. $\frac{31}{32}$
• D. $\frac{15}{16}$

Answer 1

The correct option is C $\frac{31}{32}$

By using $N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}\Rightarrow t=\frac{T_{1/2}{\log }_e\left(\frac{N_0}{N}\right)}{lo{g}_e\left(2\right)}\Rightarrow t\propto lo{g}_e\left(\frac{N_0}{N}\right)\Rightarrow \frac{t_1}{t_2}=\frac{{\left(lo{g}_e\frac{N_0}{N}\right)}_1}{{\left(lo{g}_e\frac{N_0}{N}\right)}_2}$
Hence $\frac{6}{10}=\frac{lo{g}_e\left(\frac{8}{1}\right)}{lo{g}_e\left(\frac{N_0}{N}\right)}\Rightarrow lo{g}_e\frac{N_0}{N}=\frac{10}{6}lo{g}_e\left(8\right)=lo{g}_{e}32\Rightarrow \frac{N_0}{N}=32.$
So fraction that decays =$1-\frac{1}{32}=\frac{31}{32}$.