Question

Iridium-192 is used in medicine to treat prostate cancer. Iridium-192 has two modes of radioactive decay: 96% of the time it decays by beta emission and 4% of the time it decays by electron capture. What are the daughter nuclides of these two decay processes?

Answer 1

Answer:

The daughter nuclides of these two decay processes are $^{192}_{78}Pt$ and $^{192}_{76}Os$.

Explanation:

The beta emission is represented by:

A = (Z + 1) + (n - 1) = is invariant

n: neutron

p: proton

$^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y$

Hence, the daughter nuclide of the beta emission of Ir-192 is:

$^{192}_{77}Ir \rightarrow ^{192}_{78}Pt$      

Now, electron capture is represented by:

A = (Z - 1) + (n + 1) = is invariant

$^{A}_{Z}X \rightarrow ^{A}_{Z-1}Y$

Then, the daughter nuclide of the electron capture of Ir-192 is:

$^{192}_{77}Ir \rightarrow ^{192}_{76}Os$    

Therefore, the daughter nuclides of these two decay processes are $^{192}_{78}Pt$ and $^{192}_{76}Os$.

I hope it helps you!