Respuesta :
The radioactive decay of the isotopes is an exponential decay.
In consequence, the half-life law is:
Remaining amount of isotope = Initial Quantity * [1/2]^n, where n is the number of half-lives that have elapsed.
=> Remaining amount of isotope / initial quantity = [1/2]^n.
In this case we want that the remaining quantity be 1%.
So, Remaining amount = 1% * initial quantity
=> Remaining amount / Initial quantity = 0.01
=> 0.01 = [1/2]^n
 => n log(1/2) = log(0.01)
=> n = log (0.01) / log (0.5) = 6.6439 half-lives
Now, you just must multiply the number of half-lives times the time of a half-life
=> T = 6.6439
=> n = 6.6439 * 18.27 minutes = 121.38 minutes
Answer: 121.38 minutes
In consequence, the half-life law is:
Remaining amount of isotope = Initial Quantity * [1/2]^n, where n is the number of half-lives that have elapsed.
=> Remaining amount of isotope / initial quantity = [1/2]^n.
In this case we want that the remaining quantity be 1%.
So, Remaining amount = 1% * initial quantity
=> Remaining amount / Initial quantity = 0.01
=> 0.01 = [1/2]^n
 => n log(1/2) = log(0.01)
=> n = log (0.01) / log (0.5) = 6.6439 half-lives
Now, you just must multiply the number of half-lives times the time of a half-life
=> T = 6.6439
=> n = 6.6439 * 18.27 minutes = 121.38 minutes
Answer: 121.38 minutes
For 99% of the barium-141 to be gone, 1% (0.075mg) of the starting material remains
For 7.50mg, Â
after the 1st half-life 3.75mg remain Â
after the 2nd half-life 1.875mg remain
after the 3rd half-life 0.9375mg remain
after the 4th half-life 0.46875mg remain
after the 5th half-life 0.234375mg remain
after the 6th half-life 0.1171875mg remain
after the 7th half-life 0.05859375mg remain and so on. Â
So it takes 7 half-lives to reach 99% of the barium-141 gone.
7*(18.27 minutes)= 127.89 min = 2.13 hr