Question

What percent of a sample of As-81 remains un-decayed after 43.2 seconds

Answer 1

40.4%

Step-by-step explanation:

MF=MI·(
`(1)/(2)`^n)

=100·(
`(1)/(2)`^(43.2/33))

=40.4%

Answer 2

where Ao = starting amount, At = amount after time t, and k = rate constant

You are given that the half-life is 33 seconds
Calculate the rate constant
ln(2/1) = kx33 sec
0.693 = kx33
k = 0.0210 s^-1

Now calculate the amount left after 43.2 sec
ln(100%/A1%) = 0.0210 x 43.2
ln(100/A1) = 0.9074
100/A1 = 2.478
A1 = 40.4%