So we want to know how long will it take cobalt 56 to decay to 62% to it's original value if we know that one half life is T₁/₂=79 days and it's decay constant is k=8.77*10^-3. The law for radioactive decay is:
N(t)=N₀e^-k*t, where N(t) is the quantity of the material in some time t, N₀ is the original quantity of the material, k is the decay constant, t is time.
N(t)=0.62*N₀ or 62% of the original value. 0.62*N₀=N₀e^-k*t, N₀ cancel out and we get:
0.62=e^-k*t
We need to get the time t. We get t if we take the natural logarithm of both sides:
ln(0.62)=-k*t*ln(e), ln(e)=1 so we have:
ln(0.62)=-k*t
t={ln(0.62)/(-k)} = 54.5.
Time for cobalt 56 to decay to 62% of its original value is 54.5 days.
