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An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half life of $A$ and $B$ are 10 yrs and $20 \mathrm{yrs}$ respectively. After time $t$, the alloy was found to consist of $\left(\frac{1}{e}\right) \mathrm{kg}$ of $A$ and $1 \mathrm{~kg}$ of $B$. . If the atomic weight of $A$ and $B$ are same, then the value of $t$ is (Assume, $\ln 2=0.7$ )
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The correct answer is:
$\left(\frac{200}{7}\right) \mathrm{yrs}$
Given, initial amount of radioactive element, $A=$ initial amount of radioactive element, $B$ i.e., $\left(N_0\right)_A=\left(N_0\right)_B$
half life of element $A,\left(t_2 / 2\right)_A=10 \mathrm{Yr}$
and half life of element $B,\left(t_{1 / 2}\right)_B=20 \mathrm{yr}$
After time $t$, remaining the amount of element $A$,
$N_A=\frac{1}{e} \mathrm{~kg}$
and remaining amount of element $B$,
$N_{\mathcal{B}}=1 \mathrm{~kg}$
For element $A$, after time $t$, amount of element lef undecayed is given as
$\begin{aligned}
N_A & =\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{\left(t_{1 / 2}\right)_A}} \\
\frac{1}{e} & =\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{10}}
\end{aligned}$
Similarly, for element $B$,
$\begin{gathered}
N_B=\left(N_0\right)_B\left(\frac{1}{2}\right)^{\frac{t}{20}} \\
1=\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{20}}
\end{gathered}$
Dividing Eqs. (ii) and (iii), and using Eq. (i), we get
$\frac{1}{e}=\frac{\left(\frac{1}{2}\right)^{\frac{t}{10}}}{(1)^{t / 20}}=\left(\frac{1}{2}\right)^{\frac{t}{20}}$
Taking log on the both sides, we get
$\begin{aligned}
\log \frac{1}{e} & =\frac{t}{20} \log \frac{1}{2} \\
t & =\frac{20}{\log 2} \\
& =\frac{20}{0.7}=\frac{200}{7} y \mathrm{r} \quad[\because \log 2=0.7]
\end{aligned}$
half life of element $A,\left(t_2 / 2\right)_A=10 \mathrm{Yr}$
and half life of element $B,\left(t_{1 / 2}\right)_B=20 \mathrm{yr}$
After time $t$, remaining the amount of element $A$,
$N_A=\frac{1}{e} \mathrm{~kg}$
and remaining amount of element $B$,
$N_{\mathcal{B}}=1 \mathrm{~kg}$
For element $A$, after time $t$, amount of element lef undecayed is given as
$\begin{aligned}
N_A & =\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{\left(t_{1 / 2}\right)_A}} \\
\frac{1}{e} & =\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{10}}
\end{aligned}$
Similarly, for element $B$,
$\begin{gathered}
N_B=\left(N_0\right)_B\left(\frac{1}{2}\right)^{\frac{t}{20}} \\
1=\left(N_0\right)_A\left(\frac{1}{2}\right)^{\frac{t}{20}}
\end{gathered}$
Dividing Eqs. (ii) and (iii), and using Eq. (i), we get
$\frac{1}{e}=\frac{\left(\frac{1}{2}\right)^{\frac{t}{10}}}{(1)^{t / 20}}=\left(\frac{1}{2}\right)^{\frac{t}{20}}$
Taking log on the both sides, we get
$\begin{aligned}
\log \frac{1}{e} & =\frac{t}{20} \log \frac{1}{2} \\
t & =\frac{20}{\log 2} \\
& =\frac{20}{0.7}=\frac{200}{7} y \mathrm{r} \quad[\because \log 2=0.7]
\end{aligned}$
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