k = 0.693/t1/2
Then ln(No/N) = kt
No = 50 microcuries
N = 6.25 microcuries
k from above (t1/2 = 30 years).
t = solve for this.
A Cesium 137 sample has a measured radioactivity of 50 microcuries.Cesium 137 has a half-life of 30 years.
How long will it take this sample to be reduced to a residual radioactivity of 6.25 microcuries?
Is there a formula to calculate half life?
2 answers
A 45 gram sample of a substance that's used to preserve fruit and vegetables has a k-value of 0.1109.
1 answer