Question

Derive an expression for the mean life of a radioactive substance

Answer 1

So, in order to deal this type of difficulty calculation mean life or average life of radioactive substance is introduced. ... SO the life of radioactive atoms ranges from 0-infinity mean life gives the sum of the life of all the atoms to the total no. of atoms present initially.

Answer 2

Answer:

The mean life of a radioactive substance in terms of half life time:-

Mean life period =1/decay constant = 1.44 × Half life period.

Explanation:

Consider that a radioactive element A disintegrates to give products :-

A  $\longrightarrow$  Products

Let N₀ is the atoms A present initially and N is the number of atoms present at time t. The rate of disintegration will be:-

$\frac{-dN}{dt}$  ∝ N

$\frac{-dN}{dt}= \lambda N$                                                                         .............(1)

where λ is the decay constant.

separation of variables and integrating the equation (1)

$\frac{-dN}{N} =kdt$

$\int\limits {\frac{-dN}{N} } \,=\int\limits{k} \, dt$

$-lnN=\lambda t+C$                                                                 ..................(2)

At time t=0, N = N₀ therefore, C = -ln N₀

$-lnN=\lambda t-lnN_{o}$

$lnN-lnN_{o}=- \lambda t$

$ln\frac{N_}{N_{o}}=- \lambda t$

$N=N_{0}e^{- \lambda t}$

We know that  $R=\frac{-dN}{N} =\lambda N= \lambda N_{o}e^{-\lambda t}$

$Rdt= \lambda N_{o}e^{-\lambda t} dt$

If t is life time of one nuclei then, the total time of all nuclei will be $t\lambda N_{o}e^{-\lambda t}dt$

The number of nuclei decay within the time interval 't' to  't+dt'

So, the mean life time will be:

$\tau =\lambda N_{o} \int\limits^0_{\infty} {te^{-\lambda t}} \, dt$

On solving the integral we will get

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

We know that $t_{\frac{1}{2} }= \frac{0.693}{\lambda}$

Therefore, $\tau = 1.44*t _{\frac{1}{2} }$

where, $t_{\frac{1}{2} }$ is the half life period.

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