If the half-life of an element is 69.3 text{ | vedclass

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(A) The decay constant $\lambda$ is given by $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{69.3} = 0.01 	ext{ hr}^{-1}$.Let $N_0$ be the number of

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(A) The decay constant $\lambda$ is given by $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{69.3} = 0.01 	ext{ hr}^{-1}$.Let $N_0$ be the number of active nuclei at $t = 0$.The number of active nuclei at $t = 10 	ext{ hr}$ is $N_1 = N_0 e^{-10\lambda}$.The number of active nuclei at $t = 11 	ext{ hr}$ is $N_2 = N_0 e^{-11\lambda}$.The number of nuclei that decayed between $t = 10 	ext{ hr}$ and $t = 11 	ext{ hr}$ is $\Delta N = N_1 - N_2$.The percentage of decay relative to the amount present at $t = 10 	ext{ hr}$ is:$\% 	ext{ decay} = \left( \frac{N_1 - N_2}{N_1} ight) 	imes 100$$= \left( 1 - \frac{N_2}{N_1} ight) 	imes 100$$= \left( 1 - \frac{N_0 e^{-11\lambda}}{N_0 e^{-10\lambda}} ight) 	imes 100$$= (1 - e^{-\lambda}) 	imes 100$Since $\lambda = 0.01$,$e^{-\lambda} = e^{-0.01} \approx 1 - 0.01 = 0.99$.$\% 	ext{ decay} = (1 - 0.99) 	imes 100 = 0.01 	imes 100 = 1 \%$.

If the half-life of an element is $69.3 \text{ hours}$,what percentage of it will decay between the $10^{\text{th}}$ and $11^{\text{th}}$ hours? (Initial activity $= 50 \mu Ci$)

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