Question e1cdd

Nov 7, 2016

The law of radioactive disintegration states that the rate of radioactive disintegration of a radio-element at any instant is proportional to the number of undecayed radioelement $N$ present at that instant in the sample considered.

Mathematically

$- \frac{\mathrm{dN}}{\mathrm{dt}} \propto N$
$\text{-ve sign represents decrease with time t}$

So $- \frac{\mathrm{dN}}{\mathrm{dt}} = \lambda N$

$\implies - \frac{\mathrm{dN}}{N} = \lambda \mathrm{dt}$

Where $\lambda$ is the radioactive constant

Integrating we get

$\implies - \int \frac{\mathrm{dN}}{N} = \lambda \int \mathrm{dt}$

$\implies - \ln N = \lambda t + k$

Where k = integration constant.

If at t=0 the number of undecayed atoms be ${N}_{o}$ then

$\implies - \ln {N}_{o} = \lambda d \times 0 + k$

$\implies - \ln {N}_{o} = k$

So

$\implies - \ln N = \lambda t - \ln {N}_{o}$

$\implies - \lambda t = \ln N - \ln {N}_{o} = \ln \left(\frac{N}{N} _ o\right)$

$\implies \frac{N}{N} _ o = {e}^{- \lambda t}$

$\implies N = {N}_{o} {e}^{- \lambda t}$

By radioactive disintegration law it is obvious that we cannot predict which specfic atom of the radioelement present in the sample is going to be disintegrated first. An atom may disintegrate at $t = 0 \mathmr{and} t = \infty$ So life of a radioelement is expressed as mean or average life which is defined as follows.

$\text{Aerage life} \left(\tau\right)$

$= \text{Total life of all atoms in the sample"/"Total number of atoms}$

Now

${N}_{o} \to \text{Initial number of atoms of radioelement}$

$N \to \text{No.of atoms of radioelement after t sec }$

$\mathrm{dN} \to \text{No.of atoms disintegrated in the}$
$\text{ time interval t and t+dt}$

Since dt represents infinitesimally small time then total life of dN atoms will be $t \mathrm{dN}$ and average life

$\tau = \frac{1}{N} _ o {\int}_{0}^{{N}_{o}} t \mathrm{dN}$
(As t varies from 0 to $\infty$, N varies from ${N}_{o} \to 0$ )

Now to change the variable we substitute dN obtained by differentiating $N = {N}_{o} {e}^{- \lambda t}$
So $\mathrm{dN} = - \lambda {N}_{o} {e}^{- \lambda t}$

=1/N_o int_(oo)^0t(-lambdaN_oe^(-lambdat)dt)"#

$= \frac{\lambda {N}_{o}}{N} _ o {\int}_{0}^{\infty} t {e}^{- \lambda t} \mathrm{dt}$

$= \lambda {\int}_{0}^{\infty} t {e}^{- \lambda t} \mathrm{dt}$

$= \lambda {\left[t \int {e}^{- \lambda t} \mathrm{dt} - \int \frac{d}{\mathrm{dt}} \left(t\right) \int {e}^{- \lambda t} \mathrm{dt}\right]}_{0}^{\infty}$

$= \lambda {\left[t \int {e}^{- \lambda t} \mathrm{dt} + \frac{1}{\lambda} \int {e}^{- \lambda t} \mathrm{dt}\right]}_{0}^{\infty}$

$= \lambda {\left[- \frac{t}{\lambda} {e}^{- \lambda t} - \frac{1}{\lambda} ^ 2 {e}^{- \lambda t}\right]}_{0}^{\infty}$

$= \frac{1}{\lambda}$