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)

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.