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A radioactive element gets spilled over the floor of a room. Its half-life period is $30$ days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?
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The correct answer is:
$100$ days
$100$ days
Activity $\left(-\frac{\mathrm{dN}}{\mathrm{dt}}\right) \propto \mathrm{N}$
$\mathrm{N}=\mathrm{N}_{\circ}\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\frac{\mathrm{N}}{\mathrm{N}_{\circ}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\frac{1}{10}=\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow 10=2^{\mathrm{n}}$
$\log 10=\mathrm{nlog} 2$
$\Rightarrow \mathrm{n}=\frac{1}{0.301}=3.32$
$\mathrm{t}=\mathrm{n} \times \mathrm{t}_{112}$
$=3.32 \times 30=99.6 \text { days }$
Hence, (D) is correct.
$\mathrm{N}=\mathrm{N}_{\circ}\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\frac{\mathrm{N}}{\mathrm{N}_{\circ}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$
$\frac{1}{10}=\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow 10=2^{\mathrm{n}}$
$\log 10=\mathrm{nlog} 2$
$\Rightarrow \mathrm{n}=\frac{1}{0.301}=3.32$
$\mathrm{t}=\mathrm{n} \times \mathrm{t}_{112}$
$=3.32 \times 30=99.6 \text { days }$
Hence, (D) is correct.
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