Question

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.

Answer #1

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/N_{o}) = -kt where "N" is the amount of radioisotope
remaining after time "t" has elapsed.
"N_{o}" 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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