Carbon-14 has a half-life of 5730 years. How much of a 144 g sample of carbon-14 will remain after 1719 years?

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Answer 1 · 2017-08-05T09:28:03

The mass remaining is $= 117 g$ $=117g$

The half life of carbon- $14$ $14$ is $t_{\frac{1}{2}} = 5730 y$ $t_(1/2)=5730y$

This means that in $5730 y$ $5730y$ , the initial amount of carbon $- 14$ $-14$ will be halved

The radioactive constant is

$λ = \frac{ln 2}{t_{\frac{1}{2}}} = \frac{ln 2}{5730} = 0.000121$ $lambda=ln2/t_(1/2)=ln2/5730=0.000121$

The fundamental decay equation is

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

The initial amount of carbon $- 14$ $-14$ is $m_{0} = 144 g$ $m_0=144g$

The time is $t = 1719 y$ $t=1719y$

Therefore

$m = 144 \cdot e^{- 0.000121 \cdot 1719} = 144 \cdot e^{- 0.208} = 117 g$ $m=144*e^(-0.000121*1719)=144*e^-0.208=117g$