Question

Two radioactive samples A andf B have half lives T1 and T2 (T1>T2)respectively. At t=0,the activity of B was twice the activity of A.Their activity will becomes equal after time

Answer 1

Answer:

Let decay constant be a and b respectively .

$a = \frac{ ln(2) }{t_1} \: \: and \: \: b = \frac{ ln(2) }{t_2}$

After time t both are equal

$x {e}^{ - at} = 2x {e}^{ - bt} \\ \\ - at = ( ln(2) - bt) \\ \\ (b - a)t = ln(2) \\ \\ t = \frac{ ln(2) }{b - a}$

Substitute values of a & b

$t = \frac{ ln(2) }{ \frac{ ln(2) }{t_2} - \frac{ ln(2) }{t_1} } \\\\t= \frac{t_1t_2}{t_1 - t_2}$

Answer 2

Given that,

Half lives of sample A = T₁

Half lives of sample B = T₂

At t=0,the activity of B was twice the activity of A.        

Consider initial activity of nuclei A and b be A₀ and 2A₀ respectively at a time t their activity is

For A,

$A_{A}=A_{0}e^{-\lambda t}$.....(I)

For B,

$A_{B}=2A_{A}$

$A_{B} = 2A_{0}e^{-\lambda t}$.....(II)

We need to calculate the activity after time

From equation (I) by equation (II)

$A_{A}=A_{B}$

$A_{0}e^{-\lambda_{A} t}=2A_{0}e^{-\lambda_{B} t}$

$2=\dfrac{A_{0}e^{-\lambda_{A} t}}{A_{0}e^{-\lambda_{B} t}}$

$2=\dfrac{e^{\lambda_{A} t}}{e^{\lambda_{B} t}}$

$2=e^{-\lambda_{A} t}+{\lambda_{B} t}$

$2=e^{t(\lambda_{B}-\lambda_{A})}$

Taking log on both side

$ln 2=(\lambda_{B}-\lambda_{A})t$

Put the value into the formula

$0.693=(\dfrac{0.693}{T_{2}}-\dfrac{0.693}{T_{1}})t$

$t=\dfrac{T_{2}T_{1}}{T_{1}-T_{2}}$

Hence, Their activity will becomes equal after time is $\dfrac{T_{2}T_{1}}{T_{1}-T_{2}}$