Question

The fraction of atoms of radioactive element that decays in 6 days is 7/8. The fraction that decays in 10 days will be

• A. 77/80
• B. 71/80
• C. 31/32
• D. 15/16

Answer 1

Answer: (C)

31/32

Answer 2

By using $N = N_{0}\left( \frac{1}{2} \right)^{t/T_{1/2}}$ ⇒ $t = \frac{T_{1/2}\log_{e}\left( \frac{N_{0}}{N} \right)}{\log_{e}(2)}$

⇒ $t \propto \log_{e}\frac{N_{0}}{N}$⇒ $\frac{t_{1}}{t_{2}} = \frac{\left( \log_{e}\frac{N_{0}}{N} \right)_{1}}{\left( \log_{e}\frac{N_{0}}{N} \right)_{2}}$

Hence $\frac{6}{10} = \frac{\log_{e}(8/1)}{\log_{e}(N_{0}/N)}$

⇒ $\log_{e}\frac{N_{0}}{N} = \frac{10}{6}\log_{e}(8) = \log_{e}32$⇒ $\frac{N_{0}}{N} = 32$.

So fraction that decays $= 1 - \frac{1}{32} = \frac{31}{32}$.