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A block having 12 g of element is placed in a room. This element is a radioactive element with half-life of 15 yr . After how many years will there be just 1.5 g of the element in the box?
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The correct answer is:
45 yr
$\begin{aligned} & \text { Given, } T_{1 / 2}=15 \mathrm{yr} \\ & N_0=12 \mathrm{~g} \\ & N=1.5 \mathrm{~g} \\ & \therefore \quad N=N_0 \mathrm{e}^{-\lambda t} \\ & \end{aligned}$
$\begin{array}{rlrl}\Rightarrow & & e^{-\lambda t} & =\frac{N}{N_0} \\ \Rightarrow & -\lambda t & =\ln \left(\frac{N}{N_0}\right) \\ \Rightarrow & & t & =\frac{1}{\lambda} \ln \left(\frac{N_0}{N}\right)=\frac{T_{1 / 2}}{0.693} \ln \left(\frac{N_0}{N}\right)\end{array}$
$\left[\because \lambda=\frac{0.693}{T_{1 / 2}}\right]$
$\begin{aligned} & =\frac{15}{0.693} \times \ln \left(\frac{12}{1.5}\right) \\ & =45 \mathrm{yr}\end{aligned}$
$\begin{array}{rlrl}\Rightarrow & & e^{-\lambda t} & =\frac{N}{N_0} \\ \Rightarrow & -\lambda t & =\ln \left(\frac{N}{N_0}\right) \\ \Rightarrow & & t & =\frac{1}{\lambda} \ln \left(\frac{N_0}{N}\right)=\frac{T_{1 / 2}}{0.693} \ln \left(\frac{N_0}{N}\right)\end{array}$
$\left[\because \lambda=\frac{0.693}{T_{1 / 2}}\right]$
$\begin{aligned} & =\frac{15}{0.693} \times \ln \left(\frac{12}{1.5}\right) \\ & =45 \mathrm{yr}\end{aligned}$
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