The radioisotope radon-222 has a half-life of 3.8 days. How much of a 65-g sample of radon-222 would be left after approximately 15 days?
amount left=65*e^(-.693*15/3.8)
put this in your google search window:
65*e^(-.693*15/3.8)=
To calculate the amount of a radioisotope remaining after a certain amount of time, we can use the radioactive decay formula:
N(t) = N0 * (1/2)^(t / T)
Where:
- N(t) is the remaining amount of the radioisotope after time t
- N0 is the initial amount of the radioisotope
- t is the elapsed time
- T is the half-life of the radioisotope
In this case, the initial amount is given as 65 g, the half-life is 3.8 days, and the elapsed time is approximately 15 days. We can plug these values into the formula to find the remaining amount.
N(15) = 65 * (1/2)^(15 / 3.8)
Let's calculate it step by step:
(15 / 3.8) = 3.9473684210526315
(1/2)^(3.9473684210526315) ≈ 0.067884397
Now, multiply this value by the initial amount:
N(15) ≈ 65 * 0.067884397
N(15) ≈ 4.408 spins
Therefore, approximately 4.408 g of the 65-g sample of radon-222 would be left after 15 days.