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?
A) 1.9 days
B) 23.7 days
C)54.5 days
D)79 days
E) 141 days

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? 
A) 1.9 days
B) 23.7 days
C)54.5 days
D)79 days
E) 141 days

aaronq · Answer

Use first-order integrated rate law:

$\sf \large ln[A]_t=-kt+ln[A]_o$

where $\sf ln[A]_o$ is the initial concentration of the substance
$\sf k$ is the decay constant
$\sf t$ is time
$\sf ln[A]_t$ is the concentration of substance at time $\sf t$

anonymous · Answer

The function of the half life is:$$A=A_o\cdot e^{-\lambda\cdot t}$$
$$A=100\%\cdot e^{-8.77\times10^{-3}\cdot t}$$Now we have to replace A by 62%$$62\%=100\%\cdot e^{-8.77\times10^{-3}\cdot t}$$$$0.62=e^{-8.77\times10^{-3}\cdot t}$$$$ln(0.62)=(-8.77\times10^{-3}\cdot t)\cdot ln(e)$$$$\boxed{t\approx54.5}$$

anonymous · Answer

The answer is C

anonymous · Answer

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