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How will the radioactivity of a cobalt specimen change in two years? The half-life is 5.2 years. |
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Answer» `Q=-(dN)/(dt)lamdaN`. The ratio of the activities is `x=Q_(2)/Q_(1)=N_(2)/N_(1)=E^(-lamda(t_(2)-t_(1)))=2^(-(t_(2)-t_(1))/T)` Knowing the half-life, we can easily compute the DECREASE in the activity. |
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