Question

A certain radioactive isotope has a half-life of approximately 1850 years. How many years to the nearest year would be required for a given amount of this isotope to decay to 40% of that amount?

Answer 1

N = A (0.5)^(t/j)
j = half life period = 1850 years
so amount left = 40% of original
N = 40% A
N = 0.4 A

Therefore
0.4 A = A (0.5) ^ (t/1850)

0.5 ^ (t/1850) = 0.4
log (0.5 ^ (t/1850)) = log 0.4

t/1850 = log 0.4 / log 0.5
t = (1850*log0.4) / log 0.5

Solve this,,, to get t = 2445.566 years