Question

A source contains two species of phosphorous nuclei, $${\phantom{A}}_{\phantom{15}}^{\phantom{32}}{\phantom{A}}_{\phantom{2}15}^{\phantom{2}32}\mathrm{P}$$ (T_1/2 = 14.3 d) and $${\phantom{A}}_{\phantom{15}}^{\phantom{33}}{\phantom{A}}_{\phantom{2}15}^{\phantom{2}33}\mathrm{P}$$ (T_1/2 = 25.3 d). At time t = 0, 90% of the decays are from $${\phantom{A}}_{\phantom{15}}^{\phantom{32}}{\phantom{A}}_{\phantom{2}15}^{\phantom{2}32}\mathrm{P}$$. How much time has to elapse for only 15% of the decays to be from $${\phantom{A}}_{\phantom{15}}^{\phantom{32}}{\phantom{A}}_{\phantom{2}15}^{\phantom{2}32}\mathrm{P}$$?

Answer 1

Data: $${\phantom{A}}_{\phantom{15}}^{\phantom{32}}{\phantom{A}}_{\phantom{2}15}^{\phantom{2}32}\mathrm{P}$$ : T_1/2 = 14.3 d

∴ ${\displaystyle {\lambda }_1=\frac{0.693}{14.3\text{d}}=0.04846 {\text{d}}^{-1}}$

$${\phantom{A}}_{\phantom{15}}^{\phantom{32}}{\phantom{A}}_{\phantom{2}15}^{\phantom{2}32}\mathrm{P}$$ : T_1/2 = 25.3 d

∴ ${\displaystyle {\lambda }_2=\frac{0.693}{25.3\text{d}}=0.02739 {\text{d}}^{-1}}$

At time t = 0, ${\displaystyle \frac{{\text{N}}_{\text{O1}}{\lambda }_1}{{\text{N}}_{\text{O2}}{\lambda }_2}=\frac{90\%}{10\%}=9}$  ...(1)  and

at time t, ${\displaystyle \frac{{\text{N}}_{\text{O1}}{\lambda }_1{\text{e}}^{-{\lambda }_1\text{t}}}{{\text{N}}_{\text{O2}}{\lambda }_2{\text{e}}^{-{\lambda }_2\text{t}}}=\frac{15\%}{85\%}=\frac{3}{17}}$  ...(2)

Dividing Eq. (1) by Eq. (2), we get,

${\displaystyle \frac{{\text{N}}_{\text{O1}}{\lambda }_1}{{\text{N}}_{\text{O2}}{\lambda }_2}\cdot \frac{{\text{N}}_{\text{O1}}{\lambda }_1{\text{e}}^{-{\lambda }_1\text{t}}}{{\text{N}}_{\text{O2}}{\lambda }_2{\text{e}}^{-{\lambda }_2\text{t}}}=\frac{9}{3/17}=\frac{153}{3}}$

∴ ${\displaystyle {\text{e}}^{({\lambda }_1-{\lambda }_2)\text{t}}=\frac{153}{3}}$

∴ ${\displaystyle ({\lambda }_1-{\lambda }_2)\text{t}=2.303{\log }_{10}\left(\frac{153}{3}\right)=2.303({\log }_{10}153-{\log }_{10}3)}$

∴ (0.04846 - 0.02739) t = 2.303 (2.1847 - 0.4771)

∴ t = ${\displaystyle \frac{\left(2.303\right)\left(1.7076\right)}{0.02107}}$ = 186.6 days