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Question Number 53165 by peter frank last updated on 18/Jan/19

Answered by peter frank last updated on 18/Jan/19

half life (T)  T=((0.693)/λ).......(i)  activity=(dN/dt)=Nλ.....(ii)  226g→6.023×10^(23)   1g→N  N=((6.023×10^(23) )/(226))  (dN/dt)=3.67×10^(10)

$${half}\:{life}\:\left({T}\right) \\ $$$${T}=\frac{\mathrm{0}.\mathrm{693}}{\lambda}.......\left({i}\right) \\ $$$${activity}=\frac{{dN}}{{dt}}={N}\lambda.....\left({ii}\right) \\ $$$$\mathrm{226}{g}\rightarrow\mathrm{6}.\mathrm{023}×\mathrm{10}^{\mathrm{23}} \\ $$$$\mathrm{1}{g}\rightarrow{N} \\ $$$${N}=\frac{\mathrm{6}.\mathrm{023}×\mathrm{10}^{\mathrm{23}} }{\mathrm{226}} \\ $$$$\frac{{dN}}{{dt}}=\mathrm{3}.\mathrm{67}×\mathrm{10}^{\mathrm{10}} \\ $$$$ \\ $$

Answered by peter frank last updated on 18/Jan/19

(dN/dt)=λN  T=((0.693)/λ)  N=10^8     (1) N=N_(0  ) e^(−λt)   (N/N_0 )=fraction remain=−λt  T/t=((0.693)/λ)  ....

$$\frac{{dN}}{{dt}}=\lambda{N} \\ $$$${T}=\frac{\mathrm{0}.\mathrm{693}}{\lambda} \\ $$$${N}=\mathrm{10}^{\mathrm{8}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{N}={N}_{\mathrm{0}\:\:} {e}^{−\lambda{t}} \\ $$$$\frac{{N}}{{N}_{\mathrm{0}} }={fraction}\:{remain}=−\lambda{t} \\ $$$${T}/{t}=\frac{\mathrm{0}.\mathrm{693}}{\lambda} \\ $$$$.... \\ $$

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