# If 250 mg of a radioactive element decays to 220 mg in 12 hours, how do you find the half-life of the element?

Half-life $= 65.0672 \setminus \setminus \textrm{h r s}$

#### Explanation:

The amount $N$ of a radioactive element left after time $t$ while initial amount is ${N}_{0}$ is given as

$\frac{\mathrm{dN}}{\mathrm{dt}} \setminus \propto N$

$\frac{\mathrm{dN}}{N} = - k \mathrm{dt} \setminus \quad \left(\setminus \textrm{\sin c e N \mathrm{de} c r e a s e s w . r . t . t i m e} \setminus t\right)$

$\setminus {\int}_{{N}_{0}}^{N} \frac{\mathrm{dN}}{N} = - k \setminus {\int}_{0}^{t} \mathrm{dt}$

$N = {N}_{0} {e}^{- k t}$

where, $k$ is a constant

As per given data, the ${N}_{0} = 250 \setminus m g$ of a radioactive element decays to $N = 220 \setminus m g$ in time $t = 12 \setminus \setminus \textrm{h r s}$ then we have

$220 = 250 {e}^{- k \left(12\right)}$

$k = \frac{1}{12} \setminus \ln \left(\frac{250}{220}\right)$

If ${t}_{\setminus \textrm{\frac{1}{2}}}$ is half life of radioactive element then its amount becomes ${N}_{0} / 2$ in one half line hence we have

${N}_{0} / 2 = {N}_{0} {e}^{- k \left({t}_{\setminus \textrm{\frac{1}{2}}}\right)}$

$k {t}_{\setminus \textrm{\frac{1}{2}}} = \setminus \ln 2$

setting the value of $k$ we have

$\frac{1}{12} \setminus \ln \left(\frac{250}{220}\right) {t}_{\setminus \textrm{\frac{1}{2}}} = \setminus \ln 2$

${t}_{\setminus \textrm{\frac{1}{2}}} = \setminus \frac{12 \setminus \ln 2}{\setminus \ln \left(\frac{250}{220}\right)}$

$= 65.0672 \setminus \setminus \textrm{h r s}$