24.120 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Nuclear disarmament could be accomplished if weapons were not “replenished.” The tritium in warheads decays to helium with a half-life of 12.26 yr and must be replaced or the weapon is useless. What fraction of the tritium is lost in 5.50 yr?

Step-by-Step Solution

Answer

In 5.50 years, 0.2673 percent of the tritium is lost.

Step 1: Definition of Nuclear disarmament

The act of lowering or removing nuclear weapons is known as nuclear disarmament. It might also be the final state of a nuclear-free world, in which all nuclear weapons have been eradicated. 

Step 2: Calculating fraction of the tritium

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).

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

We put N_t equal to $\frac{1}{2}{N}_0$ and solve for t_1/2 to get the half-life (t_1/2).

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

Rearranging the formula that was derived,

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

$N_t= N_0{e}^{-kt}$or $\ln \frac{N_t}{N_0} = -kt$ is an expression for calculating the number of nuclei left after a given period t.

Step 3: Calculating fraction of tritium

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

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

$k = \frac{\ln 2}{t_{\frac{1}{2}}} = \frac{\ln 2}{12.26yr} = \frac{0.05653729}{yr}$

We can solve for the value of fraction lost using the expression for calculating the number of nuclei remaining.

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

$\ln \frac{N_t}{N_0} = -\left(\frac{0.05653729}{yr}\right)(5.50yr)$

$\ln \frac{N_t}{N_0} = -0.3110$

$\frac{N_t}{N_0} = e^{-0.3110}$

$\frac{N_t}{N_0} = 0.7327$

$Fractionlost = 1-0.7327 = 0.2673$