Question

The half-life period of ${}_{53}{I}^{125}$ is 60 days. The radioactivity after 180 days will be

• A. 25%
• B. 12.5%
• C. 33.3%
• D. 3.0%

Answer 1

Answer: (B)

100% $\stackrel{\mathrm{t}{}_1/2}{\to }$50% $\stackrel{\mathrm{t}{}_1/2}{\to }$ 25% $\stackrel{\mathrm{t}{}_1/2}{\to }$ 12.5
Total time =3t1/2 = 180 days