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The Half-life of radium is 1690 years. If 70 grams are present now, how much is left in 710 years

User Danem
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Answer: 46.39 grams of radium

Explanation:

We can use the half-life formula to solve this problem:

A = A₀(1/2)^(t/t₁/₂)

where:

A₀ = initial amount (present)

A = final amount (in 710 years)

t = time elapsed (710 years)

t₁/₂ = half-life (1690 years)

First, we need to calculate the number of half-lives that will occur in 710 years:

n = t / t₁/₂

n = 710 / 1690

n ≈ 0.42

This means that in 710 years, the amount of radium will be reduced to half its current amount (1/2). And then reduced to half again (1/2 * 1/2) in another 1690 years.

Now we can calculate the final amount of radium after 710 years:

A = A₀(1/2)^n

A = 70(1/2)^0.42

A ≈ 46.39 grams

Therefore, after 710 years, approximately 46.39 grams of radium will be left.

User StephenPAdams
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