If a radioactive isotope has a half-life of 2.5 , what fraction of the original amount will remain after 10 years? How many years will it take to reduce the original amount to only 10%?
The Correct Answer and Explanation is:
To answer this question, we will use the concept of half-life, which is the time required for half of the material to decay. Here’s how we can approach each part of the question:
1. Fraction Remaining After 10 Years
The general formula for the decay of a radioactive substance is given by: N(t)=N0(12)tT1/2N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}N(t)=N0(21)T1/2t
Where:
- N(t)N(t)N(t) is the amount remaining after time ttt
- N0N_0N0 is the initial amount
- T1/2T_{1/2}T1/2 is the half-life of the isotope
- ttt is the time that has passed
For this question:
- Half-life, T1/2=2.5T_{1/2} = 2.5T1/2=2.5 years
- Time, t=10t = 10t=10 years
We substitute the given values into the equation: N(10)=N0(12)102.5=N0(12)4N(10) = N_0 \left( \frac{1}{2} \right)^{\frac{10}{2.5}} = N_0 \left( \frac{1}{2} \right)^4N(10)=N0(21)2.510=N0(21)4 N(10)=N0×116N(10) = N_0 \times \frac{1}{16}N(10)=N0×161
So, after 10 years, only 116\frac{1}{16}161 of the original amount will remain.
2. Time to Reduce the Amount to 10%
Now, we need to find the time it takes for the original amount to reduce to 10% of its initial value. We want to find the time ttt when N(t)=0.1×N0N(t) = 0.1 \times N_0N(t)=0.1×N0. Using the same decay formula: 0.1N0=N0(12)tT1/20.1 N_0 = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}0.1N0=N0(21)T1/2t
Divide both sides by N0N_0N0: 0.1=(12)t2.50.1 = \left( \frac{1}{2} \right)^{\frac{t}{2.5}}0.1=(21)2.5t
Taking the natural logarithm (ln) of both sides: ln(0.1)=t2.5ln(12)\ln(0.1) = \frac{t}{2.5} \ln\left( \frac{1}{2} \right)ln(0.1)=2.5tln(21)
Now, solve for ttt: ln(0.1)=−2.3026andln(12)=−0.6931\ln(0.1) = -2.3026 \quad \text{and} \quad \ln\left( \frac{1}{2} \right) = -0.6931ln(0.1)=−2.3026andln(21)=−0.6931 −2.3026=t2.5(−0.6931)-2.3026 = \frac{t}{2.5} (-0.6931)−2.3026=2.5t(−0.6931) t=−2.3026×2.5−0.6931=8.3 yearst = \frac{-2.3026 \times 2.5}{-0.6931} = 8.3 \, \text{years}t=−0.6931−2.3026×2.5=8.3years
Thus, it will take approximately 8.3 years to reduce the original amount to 10%.
Summary:
- After 10 years, 116\frac{1}{16}161 of the original isotope will remain.
- It will take about 8.3 years to reduce the original amount to only 10%.
