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?

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?

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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?

To find out how many half-lives of polonium-210 occur in 966 days, we need to divide the total time by the half-life of polonium-210.

Given:

Total time = 966 days
Half-life of polonium-210 = 138 days

To calculate the number of half-lives, we divide the total time by the half-life:

Number of half-lives = Total time / Half-life

Number of half-lives = 966 days / 138 days

Calculating this, we find that approximately 7. Therefore, 7 half-lives of polonium-210 occur in 966 days.

To determine how much polonium is in the sample 966 days later, we need to use the concept of exponential decay.

The exponential decay formula is:

N = N₀ * (1/2)^(t / t₁/₂)

where:
N is the final amount of polonium,
N₀ is the initial amount of polonium (given as 31 g),
t is the total time (966 days),
t₁/₂ is the half-life of polonium-210 (138 days).

Substituting these values into the formula, we have:

N = 31 g * (1/2)^(966 days / 138 days)

Calculating this, we find that approximately 0.057 g of polonium remains in the sample 966 days later.

966/138 = 7

7 half-lives means that 1/128 of the original 31g remains.