# Question #cc5db

##### 1 Answer

Here's what I got.

#### Explanation:

The **nuclear half-life**, **half** of its original mass.

If you take

#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 **undecayed** after a given period of time

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

Here **number of half-lives** that pass in the given period of time

#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 **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 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**.