# 4. Suppose a laboratory has a 26 g sample of polonium 210. The half-life of Polonium 210 about 138 days. How many half-lives of polonium 210 occur in 276 days? How much polonium 210 is in the sample 276 days later?

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

As you know, the **nuclear half-life** of a radioactive nuclide, **half** of an initial sample to decay.

This essentially means that a nuclide's half-life tells you how much time must pass in order for your sample to be reduced to half of its initial value.

In this particular case, you know that polonium-210 has a half-life of **days**, so right from the start you know that **with every** **days that pass**, the mass of the sample gets **halved**.

You can say that you have

#1/2 * A_0 = A_0/2 -># afterone half-life#1/2 * A_0/2 = A_0/4 -># aftertwo half-lives#1/2 * A_0/4 = A_0/8 -># afterthree half-lives#vdots#

and so on. The **half-life equation** can be written as

#color(blue)(ul(color(black)(A_t = A_0 * 1/2^n)))#

Here

#A_t# is the amount thatremains undecayedafter in#t# time interval#A_0# is the initial mass of the sample#n# is thenumber of half-livesthat pass in the#t# time interval

Now, notice that

#"276 days" = color(red)(2) xx "138 days"#

which means that **half-lives** pass in the given time period. Consequently, you can say that your sample will decay to

#A_"276 days" = "26 g" * 1/2^color(red)(2)#

#color(darkgreen)(ul(color(black)(A_"276 days" = "6.5 g")))#

In other words, only **days**.