If the half-life of uranium-235 is 7.04 (times) 10 (to the eight power) 8 y and 12.5 g of uranium-235 remain after 2.82 (times) 10 (to the ninth power) y, how much of the radioactive isotope was in the original sample?
To solve this problem, we can use the formula for radioactive decay:
N(t) = N₀ * (1/2)^(t / T)
Where:
- N(t) is the amount of the radioactive isotope remaining at time t
- N₀ is the initial amount of the radioactive isotope
- t is the elapsed time
- T is the half-life of the radioactive isotope
We are given:
- T (the half-life of uranium-235) = 7.04 * 10^8 years
- N(t) (amount remaining after 2.82 * 10^9 years) = 12.5 grams
Let's substitute these values into the equation:
12.5 = N₀ * (1/2)^(2.82 * 10^9 / 7.04 * 10^8)
To simplify the calculation, we can rewrite the equation as:
12.5 = N₀ * (1/2)^(4 * 10)
Now, we can solve for N₀ by isolating it on one side of the equation:
N₀ = 12.5 / (1/2)^(4 * 10)
Now, let's calculate (1/2)^(4 * 10):
(1/2)^(4 * 10) = 1/2^40 = 1 / (2^4)^10 = 1 / 16^10 = 1 / 1.09951 * 10^13
Substituting this value into the equation for N₀:
N₀ = 12.5 / (1.09951 * 10^(-13))
Now, let's calculate the value of N₀:
N₀ = 12.5 * (9.09495 * 10^12)
N₀ = 1.13686875 * 10^14 grams
Therefore, the original sample contained approximately 1.13686875 * 10^14 grams of the radioactive isotope uranium-235.