Even though plutonium$-239(t_{t/2}==2.41\times {10}^{4}yr)$is one of the main fission fuels, it is still a radiation hazard present in spent uranium fuel from nuclear power plants. How many years does it take for 99% of the plutoniumin -239 spent fuel to decay?

In physics, mass is a numerical measure of inertia, which is a fundamental feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force. The smaller the change caused by an applied force, the higher the mass of the body.

The change in the number of nuclei (N) divided by the change in time is the decay rate, or activity (A), of a radioactive sample (t).

$A=\frac{-4N}{t}$

To calculate the half-life $\left(t_{\frac{1}{2}}\right)$, we set N_t equal to$\frac{1}{2}{N}_0$ and solve for t_1/2.

$\ln \frac{N_0}{\frac{1}{2}{N}_0}=k{t}_{\frac{1}{2}}$

Rearranging the formula that was derived,

$t_{\frac{1}{2}}=\frac{\ln 2}{k}$

After a given period t, the expression for determining the number of nuclei remaining, N_t , is:

$N_t=N_0{e}^{-kt}$or$\ln \frac{N_0}{N_t}=kt$

We must use the half-life formula to find the constant (k) in this situation.

$$\begin{gathered} t_{\frac{1}{2}}=\frac{\ln 2}{k}\to k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ k=\frac{\ln 2}{t_{\frac{1}{2}}}=\frac{\ln 2}{2.41x{10}^{4}yr} \\ =\frac{2.87612938\times {10}^{-5}}{yr} \end{gathered}$$

It was mentioned in the problem that 99 percent of Pu- 239 decays. It signifies that 1% of Pu-239 is still present.

We can solve for the value of t by using the expression for calculating the number of nuclei left.

$\ln \frac{N_0}{N_t}=kt$

$$\begin{gathered} \ln \frac{1}{0.01}=\frac{2.87612938x{10}^{-5}}{yr}\left(t\right) \\ t=1.6012\times {10}^{5}yr \end{gathered}$$

Thus, the Pu-239 in spent fuel to decay in$t=1.6012\times {10}^{5}yr$