Question #1f1d8

You can't provide a definite answer here because you don't have any information about the half-life of plutonium-239, but you can work your way to an answer that depends exclusively on the half-life of the nuclide.

So, you know that radioactive decay can be modeled using the equation

#A_t = A_0 * (1/2)^(t/t_"1/2")#

Here

• #A_t# is the mass of the nuclide that remains undecayed after a given period of time #t#
• #A_0# is the initial mass of the nuclide
• #t_"1/2"# is the half-life of the nuclide

Now, in order for the nuclide to decay to #12%# of its initial mass, you must have

#A_t = 12/100 * A_0#

#A_t = 3/25 * A_0#

This basically means that after an unknown period of time #t#, the mass of plutonium-239 is reduced to #3/25# of its initial value.

Plug this back into the equation to get

#3/25 * color(red)(cancel(color(black)(A_0))) = color(red)(cancel(color(black)(A_0))) * (1/2)^(t/t_"1/2")#

#3/25 = (1/2)^(t/t_"1/2")#

Next, take the log base #10# on both sides

#log(3/25) = log[ (1/2)^(t/t_"1/2")]#

This will get you

#t/t_"1/2" * log(1/2) = log(3/25)#

Rearrange to solve for #t#

#t = log(3/25)/log(1/2) * t_"1/2"#

#t = 3.06 * t_"1/2"#

Now, the half-life of the nuclide tells you the amount of time needed for exactly half of the sample to undergo radioactive decay.

This means that you can say that your sample of plutonium-239 will decay to #12%# of its initial value after #3.06# half-lives pass.

At this point, all you have to do is to look up the half-life of plutonium-239--you can find it here--and use it to find the value of #t#. Keep in mind that you must round the answer to two sig figs.