半減期計算機

Your input: find $$$N(t)$$$ in $$$N(t)=N_0e^{-kt}$$$ given $$$N_0=250$$$, $$$t_h=15$$$, $$$t=100$$$.

$$$N(t)$$$ is the amount after the time $$$t$$$, $$$N_0$$$ is the initial amount, $$$t_h$$$ is the half-life.

First, find the constant $$$k$$$.

We know that after half-life there will be twice less the initial quantity: $$$N\left(t_h\right)=\frac{N_0}{2}=N_0e^{-k t_h}$$$.

Simplifying gives $$$\frac{1}{2}=e^{-k t_h}$$$ or $$$k=-\frac{\ln\left(\frac{1}{2}\right)}{t_h}$$$.

Plugging this into the initial equation, we obtain that $$$N(t)=N_0e^{\frac{\ln\left(\frac{1}{2}\right)}{t_h}t}$$$ or $$$N(t)=N_0\left(\frac{1}{2}\right)^{\frac{t}{t_h}}$$$.

Finally, just plug in the given values and find the unknown one.

From $$$N(t)=250\left(\frac{1}{2}\right)^{\frac{100}{15}}$$$, we have that $$$N(t)=\frac{125 \sqrt[3]{2}}{64}$$$.

Answer: $$$N(t)=\frac{125 \sqrt[3]{2}}{64}\approx 2.46078330057592$$$.