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Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half-lives of polonium-210 occur in 966 days? How much polonium is in the sample 966 days later?

Respuesta :

JcAlmighty
The answer is a) 7 half-lives and b) 0.78% (or as fraction 1/128) of the starting amount.

a) If half-life of Polonium-210 is 138 days, to calculate how many half-lives occur in 966 days, we will simply divide them: 966/138 = 7
So, 7 half-lives occur in 966 days.

b) To calculate the remaining amount, we will use the formula:
[tex] (1/2)^{n} =x[/tex]
where n is the  number of half-lives, and x is the remaining amount in decimals, and (1/2) is half-life.

We've already found that n = 7, so replace it in the formula:
[tex](1/2)^{7} =0.0078 =[/tex] 0.78% = 1/128

Answer : The amount of polonium in the sample 966 days later is, 0.24 g

Solution :

Polonium-210 is a radioactive element.

Formula used :

[tex]N(t)=N_o\times (\frac{1}{2})^{\frac{t}{t_{1/2}}[/tex]

where,

[tex]N(t)[/tex] = the amount of polonium-210 remaining after 't' days

[tex]N_o[/tex] = the initial amount of polonium-210 = 31 g

t = time = 966 days

[tex]t_{1/2}[/tex] = half-life of the polonium-210 = 138 days

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

[tex]N(t)=(31g)\times (\frac{1}{2})^{\frac{966}{138}}[/tex]

[tex]N(t)=0.24g[/tex]

Therefore, the amount of polonium in the sample 966 days later is, 0.24 g

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