Question

How much Cu-64 (half-life about 12 hours) would remain from a 2 mg sample after 12 hours?

Answer 1

Answer: 1.0 mg

Explanation: Radioactive decay follows first order kinetics.

Half-life of Cu-64 = 12 hours

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{12}=0.057hours^{-1}$

$N=N_o\times e^{-\lambda t}$

N = amount left after time t= ?

$N_0$ = initial amount  = 2 mg

$\lambda$ = rate constant  =0.057

t= time  = 12 hours

$N=2mg\times e^{- 0.057hours^{-1}\times 12hours}$

$N=1mg$

Thus amount left after 12 hours would be 1 mg as after half life, half of the initial reactant remains.

Answer 2

Answer:

1 mg

Explanation:

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