Context: I am currently writing a mathematical exploration which explores radioactive decay models in the context of the Fukushima Nuclear Incident. I am focusing specifically on two Isotopes of cesium: Cs-$134$ and Cs-$137$. I am looking at the effects it has on Game meat and Aquatic organisms (Mainly fish).
First, I obtained using Microsoft Excel a trend line for the data provided by the Japanese atomic agency. Below are the two trend lines I obtained:
One method of calculating the half-life of Cs-$134$ is to use a probability density function. The PDF gives the probability that a single Cs-$134$ particle will decay by a given time. Since the function is $N(t) = 66.851 e^{-0.0018657t}$, dividing through by $N(0)$ gives the PDF. Hence, $$\frac{N(t)}{N(0)} = \frac{66.851 e^{-0.0018657t}}{66.851} = e^{-0.0018657t}$$
However, this $t$ is in days. Converting it to years we have $$ \frac{N(t)}{N(0)} = e^{-0.0018657t \times 365} = \boxed{e^{-0.68101t}}$$
For any PDF, the area under the curve is $1$, since the sum of all outcomes is $1$. To calculate the half-life, it needs to be determined where the area under the curve is equal to $\frac{1}{2}$. So, the integral we have will be: $$\int_{0}^{a} e^{-0.68101t} \,dt = \frac{1}{2}$$
This is the way I solved it:
\begin{align*} \int_{0}^{t_{\textrm{half}}} e^{-0.68101t} \,dt &= \frac{1}{2} \\ \left[ \frac{e^{-0.68101t}}{-0.68101} \right]^{t_{\textrm{half}}}_{0} & = \frac{1}{2} \\ e^{-0.68101 t_{\textrm{half}}} - e^{0} &= \frac{1}{2} ( -0.68101) \\ a &= \frac{\ln \left[ 1 - \frac{1}{2} \left( 0.68101 \right) \right]}{-0.68101} \\ t_{\textrm{half}} &= 0.61127 \quad \textrm{(years)} \end{align*}
However, this answer is wildly different from that of other methods such as using a graphical approach:
Therefore the half-life is $1.018$, which is quite different. Another method is to use algebra:
\begin{align*} t_{\textrm{half}} &= \frac{\ln \left(\frac{1}{2} \right)}{k} \\ &= \frac{\ln \left( \frac{1}{2} \right)}{-0.68101} \\ &= 1.018 \quad \textrm{years} \end{align*}
Since the results are so different this suggests an error. However, I can only figure that it is because of $N(0)$ which is only present in the first method of using a PDF. Is there any better explanation for why this happens?
The half-life of an exponential decay process $N(t)$ with initial measured amount $N(0)$ is the time $t_{1/2}$ satisfying $$\frac{N(t_{1/2})}{N(0)} = \frac{1}{2}. \tag{1}$$ Since such a process obeys the model $$N(t) = N(0) e^{-\lambda t}, \tag{2}$$ for some rate of decay constant $\lambda > 0$, it follows that we require $$\frac{1}{2} = e^{-\lambda t_{1/2}}, \tag{3}$$ or $$t_{1/2} = \frac{\log 2}{\lambda}. \tag{4}$$ (Note that for the purposes of this discussion, all logarithms are natural.) As such, the second calculation you have indicated is the correct one.
The first calculation, which you call the "PDF" approach, is incorrect because $e^{-\lambda t}$ is not a proper density, as you can see by simply integrating:
$$\int_{t = 0}^\infty e^{-\lambda t} \, dt = \left[-\frac{1}{\lambda} e^{-\lambda t}\right]_{t=0}^\infty = 0 + \frac{1}{\lambda} e^{0} = \frac{1}{\lambda}, \tag{5}$$ which is not unity unless $\lambda = 1$. The correct density for an exponential random variable with rate parameter $\lambda$ is $$f_T(t) = \color{red}{\lambda} e^{-\lambda t}, \quad t \ge 0. \tag{6}$$
This is why your calculation does not work. Indeed, the median of such an exponential distribution is calculated as $$\frac{1}{2} = \Pr[T \le m] = \int_{t=0}^m \lambda e^{-\lambda t} \, dt = 1 - e^{-\lambda m}, \tag{7}$$ hence $$m = \frac{\log 2}{\lambda} = t_{1/2}, \tag{8}$$ the same as the half-life.
I should also take this opportunity to point out that your decay model fit should probably be described in more detail. A common approach is to perform a logarithmic transformation of the response variable (i.e., the radioactivity concentraton) and use ordinary least squares to fit a linear regression line to the log-transformed data. Specifically, this means if $t$ is time and $N(t)$ is the radioactivity, then the model used is $$\log N(t) = \beta_0 + \beta_1 t + \epsilon, \tag{9}$$ where $\beta_0$ and $\beta_1$ are the intercept and slope parameters, and $\epsilon \sim \operatorname{Normal}(0, \sigma^2)$ is a normally distributed error term with mean $0$. This also assumes the log-transformed model is homoscedastic with constant error variance and that errors are independent of $t$. Then the model transformed back to the original scale has the form $$N(t) = e^{\beta_0 + \beta_1 t + \epsilon} = e^{\beta_0} e^{\epsilon} e^{\beta_1 t}, \tag{10}$$ where now the errors are lognormally distributed. That said, this is not the only way to fit an exponential decay model to your data. It may be more illustrative for your audience to plot the data using both the linear and log-linear scales. This is easily accomplished in Excel by either changing the chart type, or by creating a new column that calculates
=ln(N). You may also wish to present the coefficient of determination $R^2$ for your model fit. The statistical aspects of your discussion and conclusions are just as important as the mathematical aspects.