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The half-life period of ${ }_{53} I^{125}$ is 60 days. The radioactivity after 180 days will be
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Verified Answer
The correct answer is:
$12.5 \%$
Let, initial concentration $(a)=100$
Given, half-life $\left(t_{1 / 2}\right)=60$ days
To find, radioactivity,
i.e. $(a-x)$ after time $T$ ( 180 days)
$$
\because \quad T=n \times t_{1 / 2}
$$
where, $n=$ no. of half-lives
$$
\begin{aligned}
180 &=n \times 60 \\
n &=\frac{180}{60}=3
\end{aligned}
$$
and $(a-x)=\frac{a}{2^{n}}=\frac{100}{2^{3}}=\frac{100}{8}$
$$
(a-x)=12.5 \%
$$
Hence, (b) is the correct answer.
Given, half-life $\left(t_{1 / 2}\right)=60$ days
To find, radioactivity,
i.e. $(a-x)$ after time $T$ ( 180 days)
$$
\because \quad T=n \times t_{1 / 2}
$$
where, $n=$ no. of half-lives
$$
\begin{aligned}
180 &=n \times 60 \\
n &=\frac{180}{60}=3
\end{aligned}
$$
and $(a-x)=\frac{a}{2^{n}}=\frac{100}{2^{3}}=\frac{100}{8}$
$$
(a-x)=12.5 \%
$$
Hence, (b) is the correct answer.
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