Question

Suppose a laboratory has a 30 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half-lives of polonium-210 occur in 1104 days? How much polonium is in the sample 1104 days later?

Answer 1

That is exactly 8 half lives.
After 8 half - lives only (1 / 2^8) or (1 / 256) of the original 30 grams will remain.
(1 / 256) = 0.003906250
So, 30 * 0.003906250 = 0.1171875 grams will remain.

Answer 2

8 half-lives of polonium-210 occur in 1104 days.

0.1174 g of polonium-210 will remain in the sample after 1104 days.

Explanation:

Initial mass of the polonium-210 = 30 g

Half life of the sample, =
`t_{(1)/(2)}=138 days`

Formula used :

`N=N_o* e^(-\lambda t)\\\\\lambda =\frac{0.693}{t_{(1)/(2)}}`

where,

`N_o` = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

`t_{(1)/(2)}` = half life of the isotope

`\lambda` = rate constant

`\lambda =(0.693)/(138 days)=0.005021 day^(-1)`

time ,t = 1104 dyas

`N=N_o* e^(-(\lambda )* t)`

Now put all the given values in this formula, we get

`N=30g* e^{-0.005021 day^(-1)* 1104 days}`

`N=0.1174 g`

Number of half-lives:

`N=(N_o)/(2^n)`

n = Number of half lives elapsed

`0.1174 g=(30 g)/(2^n)`

`n = 7.99\approx 8`

8 half-lives of polonium-210 occur in 1104 days.

0.1174 g of polonium-210 will remain in the sample after 1104 days.