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)
a. 1.9 days
b. 23.7 days
c. 54.5 days
d. 79 days
e. 141 days



Answer :

Cobalt-56 has a decay constant of 8.77 × 10-3 (which is equivalent to a half-life of 79 days).  the cobalt-56  decay time is 141 days. Option E

What is Cobalt-56?

Generally, To solve this problem, you can use the formula for exponential decay:

N(t) = N0 * e^(-kt)

where N(t) is the quantity of the substance at time t, N0 is the initial quantity of the substance, k is the decay constant, and t is the time.

We are given that the decay constant for cobalt-56 is k = 8.77 × 10^-3, and we want to find the time t it takes for the sample to decay to 62% of its original value. We can rearrange the formula to solve for t:

t = (1/k) * ln(N0/N(t))

Plugging in the values, we get:

t = (1/(8.77 × 10^-3)) * ln(1/0.62)

= 113.6 days

Therefore, the correct answer is (e) 141 days.

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