Question #7fe9c

1 Answer
Aug 12, 2015

Answer:

The half-life of the element is #t_"1/2" = "30 s"#.

Explanation:

You can actually solve this problem without using the half-life equation.

When you're dealing with a sample of a radioactive element, you know that its initial mass will be halved with every half-life that passes.

In other words, you are left with half of what you started with after one half-life, you are left with a quarter of what you starter with after two half-lives, you are left with one eighth of what you started after three half-lives, and so on.

You can actually write this as

#A = A_0/2^(n)#, where

#A# - the mass of the sample after #n# half-lives;
#A_0# - the initial mass of the sample;
#n# - the number of half-lives that have passed;

In your case, you know that you're left with #1/8"th"# of the original sample after 90 seconds. This means that you can write

#underbrace(1/8 * color(red)(cancel(color(black)(A_0))))_(color(blue)(1/8"th of the original sample")) = color(red)(cancel(color(black)(A_0))) * 1/2^n#

This is equivalent to

#1/8 = 1/2^n#

#1/2^3 = 1/2^n implies n = 3#

So, three half-lives have passed in 90 seconds, which can only mean that

#3 * t_"1/2" = "90 s" implies t_"1/2" = "90 s"/3 = color(green)("30 s")#