The decay constant of radioactive sample is $λ$ . The half-life and mean-life of the sample are, respectively, given by

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Answer 1

We know that, $N = N_{0} e^{- λ t}$
$e^{λ t} = \frac{N _{0}}{N} \Rightarrow λ t = l n (\frac{N _{0}}{N})$

For half life, $t = t_{1 / 2}, N = N_{0} / 2$ so $t_{1 / 2} = \frac{1}{λ} l n 2$

Mean life is $t_{a v g} = \frac{\int _{0}^{\infty} N t d t}{\int _{0}^{\infty} N d t}$

Now, $\int_{0}^{\infty} N t d t = \int_{0}^{\infty} N_{0} e^{- λ t} t d t = N_{0} ([(- t λ) e^{- λ t}]_{0}^{\infty} + \frac{1}{λ} \int_{0}^{\infty} e^{- λ t}) d t$ (integrated by parts )

$= 0 + \frac{N _{0}}{λ ^{2}} [e^{- λ t}]_{0}^{\infty} = \frac{N _{0}}{λ ^{2}} [0 - 1] = \frac{N _{0}}{λ ^{2}}$

$\int_{0}^{\infty} N d t = \int_{0}^{\infty} N_{0} e^{- λ t} d t = N_{0} \frac{1}{λ ^{2}} [e^{- λ t}]_{0}^{\infty} = \frac{N _{0}}{λ}$

$t_{a v g} = \frac{\frac{N _{0}}{λ ^{2}}}{N _{0} / λ} = \frac{1}{λ}$

Answer 2

We know that,

N = N_{0} e^{- λ t}

e^{λ t} = \frac{N _{0}}{N} \Rightarrow λ t = l n (\frac{N _{0}}{N})

For half life,

t = t_{1 / 2}, N = N_{0} / 2

so

t_{1 / 2} = \frac{1}{λ} l n 2

Mean life is

t_{a v g} = \frac{\int _{0}^{\infty} N t d t}{\int _{0}^{\infty} N d t}

Now,

\int_{0}^{\infty} N t d t = \int_{0}^{\infty} N_{0} e^{- λ t} t d t = N_{0} ([(- t λ) e^{- λ t}]_{0}^{\infty} + \frac{1}{λ} \int_{0}^{\infty} e^{- λ t}) d t

(integrated by parts )

= 0 + \frac{N _{0}}{λ ^{2}} [e^{- λ t}]_{0}^{\infty} = \frac{N _{0}}{λ ^{2}} [0 - 1] = \frac{N _{0}}{λ ^{2}}

\int_{0}^{\infty} N d t = \int_{0}^{\infty} N_{0} e^{- λ t} d t = N_{0} \frac{1}{λ ^{2}} [e^{- λ t}]_{0}^{\infty} = \frac{N _{0}}{λ}

t_{a v g} = \frac{\frac{N _{0}}{λ ^{2}}}{N _{0} / λ} = \frac{1}{λ}