Question #cc5db

The nuclear half-life, #t_"1/2"#, tells you the amount of time needed in order for an initial sample of a radioactive nuclide to decay to half of its original mass.

If you take #B_0# to be the initial mass of the radioactive substance, you can say that you will have

• #B_0 * 1/2 = B_0 * (1/2)^color(red)(1)-># after #color(red)(1)# half-life
• #B_0/2 * 1/2 = B_0 * 1/4 = B_0 * (1/2)^color(red)(2) -># after #color(red)(2)# half-lives
• #B_0/4 * 1/2 = B_0 * 1/8 = B_0 * (1/2)^color(red)(3) -># after #color(red)(3)# half-lives
• #B_0/8 * 1/2 = B_0 * 1/16 = B_0 * (1/2)^color(red)(4) -># after #color(red)(4)# half-lives
  #vdots#

and so on. Now, if you take #B_t# to be the amount that remains undecayed after a given period of time #t#, you can say that #B_t# will be equal to

#B_t = B_0 * (1/2)^color(red)(n)#

Here #color(red)(n)# represents the number of half-lives that pass in the given period of time #t#. You can express the number of half-lives that pass in the given period of time #t# by using the half-life of the substance

#color(red)(n) = t/t_"1/2" color(white)((acolor(black)( larr " the total time that passes")/(color(black)( larr " one half-life")aaaaaaaaaaa)#

In your case, you have

#t_"1/2" = "40 years" " "# and #" " B_0 = "3 g"#

so

#color(red)(n) = t/"40 years"#

and the exponential equation will look like this

#color(darkgreen)(bar(|ul(color(black)(color(white)(a/a)B_t = "3 g" * (1/2)^(t/"40 years")color(white)(a/a)))|))#

To determine how much #B# is left after #8# years, simply calculate the number of half-lives that will pass in this amount of time

#(8 color(red)(cancel(color(black)("years"))))/(40color(red)(cancel(color(black)("years")))) = 1/5#

and plug the result into the equation

#B_"8 years" = "3 g" * (1/2)^(1/5) = color(darkgreen)(ul(color(black)("2.61 g B")))#

Finally, to determine the number of years needed for the initial sample to be reduced to #"0.5 g"#, rearrange the equation as

#0.5 color(red)(cancel(color(black)("g"))) = 3 color(red)(cancel(color(black)("g"))) * (1/2)^(t/"40 years")#

#(1/2)^(t/"40 years") = 0.5/3#

Take the natural log from both sides

#ln[(1/2)^(t/"40 years")] = ln(1/6)#

This will be equivalent to

#t/"40 years" * ln(1/2) = ln(1/6)#

You will end up with

#t = ln(1/6)/ln(1/2) * "40 years" = color(darkgreen)(ul(color(black)("103 years")))#

I'll leave the values rounded to three sig figs.