Which sample, $A$ or $B$ shown in figure has shorter mean-life?

Question

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Answer 1

Initially at $t = 0$ from figure given
$(\frac{d N _{0}}{d t})_{A} = (\frac{d N _{0}}{d t})_{B}$
So $(N_{0})_{A} = (N_{0})_{B}$
ie initially both samples has equal number of radioactive atoms. Considering at any instant $t = t$ from figure.
$(\frac{d N _{0}}{d t})_{A} > (\frac{d N _{0}}{d t})_{B}$
$λ_{A} N_{A} = λ_{B} N_{B}$
$N_{A} > N_{B}$
$λ_{A} < λ_{B} (\frac{- d N}{d t} = λ N)$
$∴ τ = \frac{1}{λ}$
$∴ \frac{1}{τ _{A}} < \frac{1}{τ _{B}}$
$τ_{A} > τ_{B}$
So mean life time of sample $A$ is greater than of $B$

Answer 2

Initially at

t = 0

from figure given

(\frac{d N _{0}}{d t})_{A} = (\frac{d N _{0}}{d t})_{B}

So

(N_{0})_{A} = (N_{0})_{B}

ie initially both samples has equal number of radioactive atoms. Considering at any instant

t = t

from figure.

(\frac{d N _{0}}{d t})_{A} > (\frac{d N _{0}}{d t})_{B}

λ_{A} N_{A} = λ_{B} N_{B}

N_{A} > N_{B}

λ_{A} < λ_{B} (\frac{- d N}{d t} = λ N)

∴ τ = \frac{1}{λ}

∴ \frac{1}{τ _{A}} < \frac{1}{τ _{B}}

τ_{A} > τ_{B}

So mean life time of sample

A

is greater than of

B