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Suppose a laboratory has a 38 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?
9 ; 0.07 g
8 ; 2,2622 g
8 ; 4.75 g
8 ; 0.15 g

Respuesta :

The answer is the last one which is "8 ; 0.15 g". Please see below solution:

mo x (1/2)^t/t 1/2 = mt
38 x (1/2)^1104/138 = mt 
mt = 38 (1/2)^8
mt = 38/256
mt = 0.1484 
then round up the answer 
then it will 0.15 g
InesWalston

Answer:

0.15 g of Polonium will be in the sample 1104 days later.

Step-by-step explanation:

The exponential function for decay is,

[tex]y(t)=a(1-r)^t[/tex]

Where,

y(t) = the amount after time t,

a = initial amount = 38 g

r = rate of decay = 50% = 0.50 (as the sample is getting halved each time)

t = time period = [tex]\dfrac{1104}{138}=8[/tex] (as we have to convert the time in terms of half lifes)

Putting all the values,

[tex]y(t)=38(1-0.5)^8[/tex]

[tex]=38(0.5)^8[/tex]

[tex]=0.15\ g[/tex]

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