24.126 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Definition of change of mass

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.

Step 2: Find how many years does it take for 99 % of the plutonium -239 in spent fuel to decay?

 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 $\ln \frac{N_0}{N_t}=kt$

Step 3:Use half life formula

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 $t=1.6012\times {10}^{5}yr$