Vraag 23

uitkomst:$m=2{,}6 \cdot 10^{-5} \mathrm{~kg}

voorbeeld van een berekening:

Het aantal atoomkernen kan berekend worden met behulp van de activiteit:

$A=\frac{\ln 2}{t_{\frac{1}{2}}} N \rightarrow N=\frac{A t_{\frac{1}{2}}}{\ln 2}.

Voor de halveringstijd van cobalt-60 geldt: t_{\frac{1}{2}}=5{,}27\,\mathrm{jr}=1{,}66\cdot10^8\mathrm{~s}t_{\frac{1}{2}}=5{,}27\mathrm{jr}=1{,}66\cdot10^8\mathrm{~s}t_{\frac{1}{2}}=5{,}27\mathrm{jr}=1{,}66\cdot10^8\mathrm{~s}.

Voor het aantal atoomkernen geldt daarmee: N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2{,}64\cdot10^{20}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2{,}64\cdot10^2N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2{,}64\cdot10N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2{,}64\cdot1N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2{,}64\cdotN=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2{,}64N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2{,}6N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2{,}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=2N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}=N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln2}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\ln}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{l}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10^8}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot10}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot1}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66\cdot}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}66}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}6}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1{,}}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot1}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}\cdot}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^{12}}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^1}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^12}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10^1}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot10}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot1}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1\cdot}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}1}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1{,}}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1.}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{1}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{\placeholder{}}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=1\frac{\placeholder{}}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=\frac{\placeholder{}}{\placeholder{}}N=\frac{At_{\frac12}}{\ln2}=N=\frac{At_{\frac12}}{\ln2}N=\frac{At_{\frac12}}{\ln}N=\frac{At_{\frac12}}{\ln\left(\right)}N=\frac{At_{\frac12}}{\ln\left(2\right)}N=\frac{At_{\frac12}}{\ln\left(\right)}N=\frac{At_{\frac12}}{\ln}N=\frac{At_{\frac12}}{l}N=\frac{At_{\frac12}}{\placeholder{}}N=At_{\frac12}N=At_{\frac12\frac{\placeholder{}}{\placeholder{}}}N=At_{\frac12}N=At_{\frac{1}{\placeholder{}}}N=At_1N=AtN=At-N=AtN=AN=N.

De massa van het cobalt-60 in een bron kan berekend worden met de atoommassa:

m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}6\cdot10^{-5}\operatorname{\mathrm{kg}}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}6\cdot10^{-5}km=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}6\cdot10^{-5}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}6\cdot10^{-}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}6\cdot10m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}6\cdot1m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}6\cdotm=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}6m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2{,}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=2m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}=m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}-m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-27}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-2}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10^{-}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot10m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdot1m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66\cdotm=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}66m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}6m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1{,}m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdot1m=Nm_{at}=2{,}64\cdot10^{20}\cdot60\cdotm=Nm_{at}=2{,}64\cdot10^{20}\cdot60m=Nm_{at}=2{,}64\cdot10^{20}\cdot6m=Nm_{at}=2{,}64\cdot10^{20}\cdotm=Nm_{at}=2{,}64\cdot10^{20}m=Nm_{at}=2{,}64\cdot10^2m=Nm_{at}=2{,}64\cdot10m=Nm_{at}=2{,}64\cdot1m=Nm_{at}=2{,}64\cdotm=Nm_{at}=2{,}64m=Nm_{at}=2{,}6m=Nm_{at}=2{,}m=Nm_{at}=2m=Nm_{at}=m=Nm_{at}m=Nm_{a}m=Nmm=Nm=m.