If half life of an element is 69.3 hours then | vedclass

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Question

Consider active Nuclei at $t =0$ is $N_{0}$ then,

The active nuclei at $t =10 hr$

$N_{1}=N_{0} e^{-10 \lambda}$

The active nuclei at $t=11 hr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

Consider active Nuclei at $t =0$ is $N_{0}$ then,

The active nuclei at $t =10 hr$

$N_{1}=N_{0} e^{-10 \lambda}$

The active nuclei at $t=11 hr$

$N_{2}=N_{0} e^{-11 \lambda}$

The expression for the decay percentage,

$\%$ decay $=\left(\frac{N_{1}-N_{2}}{N_{1}}ight) 	imes 100$

$=\left(\frac{N_{0} e^{-10 \lambda}-N_{0} e^{-11 \lambda}}{N_{0} e^{-10 \lambda}}ight)$

$=\left(1-\frac{1}{e^{\lambda}}ight) 	imes 100$

$=\left(1-\frac{1}{e^{\ln 2 / T_{1 / 2}}}ight) 	imes 100 	imes 100$

The half life of the element is, $T_{1 / 2}=69.3 hr$ therefore,

$\%$ decay $=\left(1-\frac{1}{e^{0.01}}ight) 	imes 100$

$=1 \%$

If half life of an element is $69.3$ hours then how much of its percent will decay in $10^{\text {th }}$ to $11^{\text {th }}$ hours. Initial activity $=50\, \mu Ci$

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