bortefterxq5l

2023-02-26

The half-life for the radioactive decay of C-14 is 5730 years.How long will it take for 30% of the C-14 atoms in a sample of C-14 to decay?

Izaiah Bean

Beginner2023-02-27Added 6 answers

${t}_{\frac{1}{2}}=5730y$

$\lambda =\frac{\mathrm{ln}2}{{t}_{\frac{1}{2}}}=\frac{\mathrm{ln}2}{5730}=1.21\cdot {10}^{-4}{y}^{-1}$

$m\left(t\right)={m}_{0}{e}^{\lambda t}$

$\frac{m\left(t\right)}{{m}_{0}}={e}^{\lambda t}$

$\lambda t=\mathrm{ln}\left(\frac{m\left(t\right)}{{m}_{0}}\right)$

$t=\frac{1}{\lambda}\mathrm{ln}\left(\frac{m\left(t\right)}{{m}_{0}}\right)$

$t=\frac{1}{1.21\cdot {10}^{-4}}\cdot \mathrm{ln}\left(0.7\right)=2948.5y$

$\lambda =\frac{\mathrm{ln}2}{{t}_{\frac{1}{2}}}=\frac{\mathrm{ln}2}{5730}=1.21\cdot {10}^{-4}{y}^{-1}$

$m\left(t\right)={m}_{0}{e}^{\lambda t}$

$\frac{m\left(t\right)}{{m}_{0}}={e}^{\lambda t}$

$\lambda t=\mathrm{ln}\left(\frac{m\left(t\right)}{{m}_{0}}\right)$

$t=\frac{1}{\lambda}\mathrm{ln}\left(\frac{m\left(t\right)}{{m}_{0}}\right)$

$t=\frac{1}{1.21\cdot {10}^{-4}}\cdot \mathrm{ln}\left(0.7\right)=2948.5y$