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

Answer:

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

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

Step-by-step explanation:

Initial mass of the polonium-210 = 30 g

Half life of the sample, = $t_{\frac{1}{2}}=138 days$

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

where,

$N_o$ = initial mass of isotope

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

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$\lambda =\frac{0.693}{138 days}=0.005021 day^{-1}$

time ,t = 1104 dyas

$N=N_o\times e^{-(\lambda )\times t}$

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

$N=30g\times e^{-0.005021 day^{-1}\times 1104 days}$

$N=0.1174 g$

Number of half-lives:

$N=\frac{N_o}{2^n}$

n =  Number of half lives elapsed

$0.1174 g=\frac{30 g}{2^n}$

$n = 7.99\approx 8$

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

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