The half-life period of `I_(53)^(125)` is 60 days. What percent of the original radioactivity will be present after 180 days ?

Question

The half-life period of `I_(53)^(125)` is 60 days. What percent of the original radioactivity will be present after 180 days ?

The half-life period of `I_(53)^(125)` is 60 days. What percent of the original radioactivity will be present after 180 days ?
A. `50%`
B. `20.5%`
C. `12.5%`
D. `25%`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
`t_(1//2)=60 "days", " and " t=180 "days"`
Therefore, `t//t_(1//2) = (180)/(60)=3.0 " Or "N=(N_(0))/(2^(3.0))=(1)/(8)N_(0)=(1)/(8)xx100` Precent of `N_(0)=12.5%`

4 views Answered 2023-02-05 15:29