Answer to Question #301090 in Chemistry for Kimmy

Question

Au – 198 is used in the diagnosis of liver problem. The half- life of Au -198 is 2.69 days.

If you begin with 2.8 µg of this gold isotope, what mass remains in 10.8 days?

Answer 1

Solution:

The half-life (t_1/2) of Au-198 is 2.69 days.

From law of radioactive decay:

t_1/2 = ln(2) / λ

where λ = decay constant

Therefore,

λ = ln(2) / t_1/2

λ = (0.693) / (2.69 days) = 0.2576 days⁻¹

Radioactive decay follows the kinetics of a first order reaction:

ln[Au-198]_t = −λt + ln[Au-198]_o

[Au-198]_t = unknown

[Au-198]_o = 2.8 µg

λ = 0.2576 days⁻¹

t = 10.8 days

Therefore,

ln[Au-198]_t = −(0.2576 days⁻¹ × 10.8 days) + ln(2.8)

ln[Au-198]_t = −2.78208 + 1.02962

ln[Au-198]_t = −1.75246

[Au-198]_t = e^−1.75246 = 0.1733 µg = 0.17 µg

[Au-198]_t = 0.17 µg

Answer: 0.17 µg of Au-198 remains