+29
What is the half-life (in days) if the decay each day is 2% ?
Also how can I do this?
+26393 What is the half-life (in days) if the decay each day is 2% ?
Also how can I do this
\(\begin{array}{rcl} y &=& a \cdot b^t \qquad | \qquad y = \frac{a}{2} \\ \frac{a}{2}&=& a \cdot b^{\frac{h}{1~\text{day}}} \qquad | \qquad h = \text{half-life} \\ \frac{1}{2}&=& b^h \qquad | \qquad \log{} \\ \log{ (\frac{1}{2}) } &=& h\cdot \log{(b)} \\ h &=& \frac{ \log{ (\frac{1}{2}) } } { \log{(b)} } \qquad | \qquad 2~\% \rightarrow 1-\frac{p}{100}=1-\frac{2}{100} = 0.98= b\\ h &=& \frac{ \log{ (\frac{1}{2}) } } { \log{(0.98)} } \\ h &=& \frac{ -0.30102999566 } {-0.00877392431} \\ h &=& 34.3096184915 ~\text{days} \\ \end{array}\)
The half-life is 34.3 days
+26393 What is the half-life (in days) if the decay each day is 2% ?
Also how can I do this
\(\begin{array}{rcl} y &=& a \cdot b^t \qquad | \qquad y = \frac{a}{2} \\ \frac{a}{2}&=& a \cdot b^{\frac{h}{1~\text{day}}} \qquad | \qquad h = \text{half-life} \\ \frac{1}{2}&=& b^h \qquad | \qquad \log{} \\ \log{ (\frac{1}{2}) } &=& h\cdot \log{(b)} \\ h &=& \frac{ \log{ (\frac{1}{2}) } } { \log{(b)} } \qquad | \qquad 2~\% \rightarrow 1-\frac{p}{100}=1-\frac{2}{100} = 0.98= b\\ h &=& \frac{ \log{ (\frac{1}{2}) } } { \log{(0.98)} } \\ h &=& \frac{ -0.30102999566 } {-0.00877392431} \\ h &=& 34.3096184915 ~\text{days} \\ \end{array}\)
The half-life is 34.3 days
+33661 Another way of looking at this is as follows:
If you start with N0, then after 1 day you are left with 0.98*N0
After 2 days you are left with 0.98^2*N0
After 3 days you are left with 0.98^3*N0
...
After t days you are left with 0.98^t*N0
For t to be the half-life you must have 0.98^t*N0 = N0/2
Divide by N0 and take logs of both sides:
ln(0.98^t) = ln(1/2)
t*ln(0.98) = ln(1/2)
t = ln(1/2)/ln(0.98) ≈ 34.3 days
