Question

# Radioactive gold-198 is used in the diagnosis of liver problems. 198Au decays in a first-order process,...

Radioactive gold-198 is used in the diagnosis of liver problems. 198Au decays in a first-order process, emitting a β particle (electron). The half-life of this isotope is 2.7 days. You begin with a 5.6-mg sample of the isotope. Calculate how much gold-198 remains after 3.5 days.

Step 1: Half life of a first order process is

t1/2 = 0.693/k where k is rate constant and t1/2 is half life.

Here t1/2 is given, t1/2 = 2.7 days. So the rate constant of the reaction is

k= 0.693/2.7 = 0.26 per day. So,the rate constant of the reaction is 0.26/day.

ln(N/No) = -kt   where "N" is the amount of radioisotope remaining after time "t" has elapsed. "No" is the initial amount of radioisotope at the beginning of the period, and "k" is the rate constant for the radioisotope being studied. In this equation, the units of measure for N and No can be in grams, atoms, or moles.

How much of a 5.6 mg sample of isotope (k = 0.26 day-1) will remain after 3.5 days?

ln(N/5.6 mg) = -(0.26 day-1)(3.5 days)
ln(N/5.6 mg) = -0.91
N/5.6 mg = e-0.91 = 0.09
N = 0.504 mg

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