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?

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Answer 1 · 2017-12-17T08:13:46

The time is $= 2948.5 y$ $=2948.5y$

The half life of $C - 14$ $C-14$ is

$t_{\frac{1}{2}} = 5730 y$ $t_(1/2)=5730y$

The radioactive constant is

$λ = \frac{ln 2}{t_{\frac{1}{2}}} = \frac{ln 2}{5730} = 1.21 \cdot 10^{- 4} y^{- 1}$ $lambda=ln2/t_(1/2)=ln2/5730=1.21*10^-4y^-1$

Apply the equation

$m (t) = m_{0} e^{λ t}$ $m(t)=m_0e^(lambdat)$

$\frac{m (t)}{m_{0}} = e^{λ t}$ $(m(t))/m_0=e^(lambdat)$

$λ t = ln (\frac{m (t)}{m_{0}})$ $lambdat=ln((m(t))/m_0)$

$t = \frac{1}{λ} ln (\frac{m (t)}{m_{0}})$ $t=1/lambdaln((m(t))/m_0)$

The time is

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