A radioactive isotope has a half-life of 2.5 years. Which fraction of the original mass remains unchanged after 10 years?
The Correct Answer and Explanation is:
To find the fraction of the original mass of a radioactive isotope that remains unchanged after 10 years, we can use the concept of half-life and the following formula:N(t)N0=(12)tT1/2\frac{N(t)}{N_0} = \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}N0N(t)=(21)T1/2t
Where:
- N(t)N(t)N(t) is the amount of the isotope remaining after time ttt.
- N0N_0N0 is the initial amount of the isotope.
- T1/2T_{1/2}T1/2 is the half-life of the isotope.
- ttt is the time that has passed (in years).
Given:
- Half-life T1/2=2.5T_{1/2} = 2.5T1/2=2.5 years.
- Time t=10t = 10t=10 years.
Plugging these values into the equation:N(t)N0=(12)102.5=(12)4\frac{N(t)}{N_0} = \left( \frac{1}{2} \right)^{\frac{10}{2.5}} = \left( \frac{1}{2} \right)^4N0N(t)=(21)2.510=(21)4N(t)N0=116\frac{N(t)}{N_0} = \frac{1}{16}N0N(t)=161
Thus, after 10 years, only 116\frac{1}{16}161 of the original amount of the isotope remains.
Explanation:
The radioactive decay process follows an exponential decay model, where the substance loses half of its remaining mass after each half-life. In this case, the half-life is 2.5 years, meaning that after each 2.5-year period, half of the isotope’s mass is gone.
After 10 years, the number of half-lives that have passed is:102.5=4 half-lives\frac{10}{2.5} = 4 \text{ half-lives}2.510=4 half-lives
Each half-life reduces the remaining amount by half, so after four half-lives, the fraction of the isotope remaining is (12)4=116\left( \frac{1}{2} \right)^4 = \frac{1}{16}(21)4=161. This means only 1/16 of the original mass of the isotope is still present after 10 years.
