At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio $\frac{R _{B}}{R _{A}}$ of their activities after time t itself decays with time t as $e^{- 3 t} .$ If the half-life of A is $l n 2$ , the half-life of B is:

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Answer 1

Half life of $A = l n 2$
$t_{1 / 2} = \frac{ℓ n 2}{λ}$
$λ_{A} = 1$
at $t = 0 R_{A} = R_{B}$
$N_{A} e^{- λ A T} = N_{B} e^{- λ B T}$
$N_{A} = N_{B}$ at t = 0
at t= t $\frac{R _{B}}{R _{A}} = \frac{N _{0} e ^{- λ_{B}^{t}}}{N _{0} e ^{- λ_{A}^{t}}}$
$e^{- (λ_{B} - λ_{A}) t} = e^{- 3 t}$
$λ_{B} - λ_{A} = 3$
$λ_{B} = 3 + λ_{A} = 4$
$t_{1 / 2} = \frac{ℓ n 2}{λ _{B}} = \frac{ℓ n 2}{4}$

Answer 2

Half life of

A = l n 2

t_{1 / 2} = \frac{ℓ n 2}{λ}

λ_{A} = 1

at

t = 0 R_{A} = R_{B}

N_{A} e^{- λ A T} = N_{B} e^{- λ B T}

N_{A} = N_{B}

at t = 0
at t= t

\frac{R _{B}}{R _{A}} = \frac{N _{0} e ^{- λ_{B}^{t}}}{N _{0} e ^{- λ_{A}^{t}}}

e^{- (λ_{B} - λ_{A}) t} = e^{- 3 t}

λ_{B} - λ_{A} = 3

λ_{B} = 3 + λ_{A} = 4

t_{1 / 2} = \frac{ℓ n 2}{λ _{B}} = \frac{ℓ n 2}{4}