Question

Cobalt-56 has a decay constant of 8.77 × 10-3 (which is equivalent to a half life of 79 days). How many days will it take for a sample of cobalt-56 to decay to 62% of its original value? (At = A0e-kt)

Answer 1

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.