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Write the definition of half life of radioactive substance and obtain its relation to decay constant.
Solution
"The time interval during which the number of nuclei of radioactive element reduces to half its value at the beginning of the interval is called half life".
$\therefore$ Half life $\left(\mathrm{T}_{1 / 2}\right)=$ the number of nuclei of beginning decreases in half.
$=\frac{\mathrm{N}_{0}}{2}$
but in exponential law $\mathrm{N}=\mathrm{N}_{0} e^{-\lambda t}, \mathrm{~N}=\frac{\mathrm{N}_{0}}{2}$ and putting $t=\mathrm{T}_{1 / 2}$,
$\frac{\mathrm{N}_{0}}{2}=\mathrm{N}_{0} e^{-\lambda \mathrm{T}_{1} / 2}$ $\frac{1}{2}=e^{-\lambda \mathrm{T}_{1 / 2}}$
$\therefore 2=e^{\lambda \mathrm{T}_{1 / 2}}$ Taking $\log$ on both sides, $\therefore \ln 2=\lambda \mathrm{T}_{1 / 2} \cdot \ln e$
$\therefore \log _{e} 2=\lambda \mathrm{T}_{1 / 2} \cdot \log _{e} e$
$\therefore 2.303 \times \log _{10} 2=\lambda \mathrm{T}_{1 / 2} \times 1 \quad\left[\because \log _{e} e=1\right]$
$\therefore 2.303 \times 0.3010=\lambda \mathrm{T}_{1 / 2}$
$\therefore 0.693=\lambda \mathrm{T}_{1 / 2}$
$\therefore \mathrm{T}_{1 / 2}=\frac{0.693}{\lambda}$
Hence, half life of a radioactive element is inversely proportional to the decay constant and independent to the number of nucleus present in the sample.
Disintegration rate also defines half life. "The period in which the radioactive sample is half of the initial activity $\left(\mathrm{R}_{0}\right)$ is called its half life".
