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Ba-122 has half life of 2 min. Experiment has to be done using Ba-122 and it takes 10 min to set up the experiment. If initially 80 g of Ba-122 was taken, how much Ba was left when experiment was started?
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The correct answer is:
2.5 g
As $\mathrm{T}_{\frac{1}{2}}=2 \mathrm{~min} \Rightarrow 10 \mathrm{~min}=5$ half-lives
We know, $\frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}$
where, $\quad \mathrm{N}=$ remaining of nuclei
$\mathrm{N}_0=$ initial number of nuclei
$\mathrm{n}=$ number of half lives
$\Rightarrow \quad \mathrm{N}=\frac{\mathrm{N}_0}{2^n}=\frac{80}{2^5}=\frac{80}{32}=2.5 \mathrm{~g}$
We know, $\frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}$
where, $\quad \mathrm{N}=$ remaining of nuclei
$\mathrm{N}_0=$ initial number of nuclei
$\mathrm{n}=$ number of half lives
$\Rightarrow \quad \mathrm{N}=\frac{\mathrm{N}_0}{2^n}=\frac{80}{2^5}=\frac{80}{32}=2.5 \mathrm{~g}$
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