The half-life of a radioactive nuclide is 40 | vedclass

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(B) The fraction of the substance remaining undecayed is given by $\frac{N}{N_0} = 2^{-n}$, where $n = \frac{t}{T_{1/2}}$.Given $t = 20 \, 	ext{hours

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$\frac{\sqrt{2} - 1}{\sqrt{2}}$

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(B) The fraction of the substance remaining undecayed is given by $\frac{N}{N_0} = 2^{-n}$, where $n = \frac{t}{T_{1/2}}$.Given $t = 20 \, 	ext{hours}$ and $T_{1/2} = 40 \, 	ext{hours}$, the number of half-lives $n = \frac{20}{40} = 0.5$.Therefore, the fraction remaining is $\frac{N}{N_0} = 2^{-0.5} = \frac{1}{2^{0.5}} = \frac{1}{\sqrt{2}}$.The fraction that has decayed is $1 - \frac{N}{N_0}$.Decayed fraction $= 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$.

The half-life of a radioactive nuclide is $40 \, \text{hours}$. Find the fraction of the substance that has decayed after $20 \, \text{hours}$.

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