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Question Number 116238 by Ar Brandon last updated on 02/Oct/20

$$\mathrm{How}\mathrm{long}\mathrm{does}\mathrm{it}\mathrm{take}\mathrm{for}\mathrm{the}\mathrm{disintegration}\mathrm{of}$$$$\frac{3}{10}\mathrm{of}\mathrm{an}\mathrm{atom}\mathrm{with}\mathrm{radioactive}\mathrm{constant}\lambda ?$$

Answered by MrGaster last updated on 22/Jan/25

$$N\left(t\right)=N_0{e}^{-\lambda t}$$$$\frac{3}{10}{N}_0=N_0{e}^{-\lambda t}$$$$\frac{3}{10}=e^{-\lambda t}$$$$\ln \left(\frac{3}{10}\right)=-\lambda t$$$$t=-\frac{\ln \left(\frac{3}{10}\right)}{\lambda }$$$$t=\frac{\ln \left(\frac{10}{3}\right)}{\lambda }$$

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