Question #7f902

The idea here is that the amount of radium-226 you have in your sample will be halved with every passing half-life.

This is the case because a radioactive nuclide's nuclear half-life tells you the amount of time needed for half of an initial sample to undergo radioactive decay.

Mathematically, you can write this as

#A_t = A_0 * (1/2)^color(red)(n)#

Here

• #A_t# is the amount of the radioactive nuclide that remains undecayed after a period of time #t#
• #A_0# is the initial amount of the radioactive nuclide
• #color(red)(n)# is the number of half-lives that pass in the period of time #t#

Now, you know that #1/8"th"# of an initial sample of radium-226 remains undecayed. This is equivalent to saying that you have

#A_t = A_0 * 1/8#

Since you know that

#8 = 2 * 2 * 2 = 2^3#

you can rewrite this as

#A_t = A_0 * (1/2)^color(red)(3)#

You can thus say that in order for the initial sample to be reduced to #1/8"th"# of its initial value, #color(red)(n) = color(red)(3)# half-lives must pass.

Since you know that radium-226 has a half-life of #1599# years, you can say that a total of

#3 color(red)(cancel(color(black)("half-lives"))) * "1599 years"/(1color(red)(cancel(color(black)("half-life")))) = color(darkgreen)(ul(color(black)("4797 years")))#

The answer is rounded to four sig figs, the number of sig figs you have for the half-life of the nuclide.

Keep in mind that you have three sig figs for the initial mass of the sample, but you're not using this value in your calculations.