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Chapter 3: Problem 5

Suppose a radioactive isotope is such that one-fifth of the atoms in a sample decay after three years. Find the half-life of this isotope.

Short Answer

Expert verified
The half-life of the radioactive isotope is approximately 7.42 years.

Step by step solution

01

Fill in the known values in the exponential decay formula

We know that after three years, four-fifths of the atoms will remain, which means \(N(3) = 4/5 N_0\). Now we can plug this value and t = 3 into the decay formula: \(\frac{4}{5}N_0 = N_0 e^{-3\lambda}\)
02

Solve for the decay constant 位

Divide both sides of the equation by \(N_0\) to get: \(\frac{4}{5} = e^{-3\lambda}\) Now take the natural logarithm of both sides to solve for the decay constant \(\lambda\): \(ln(\frac{4}{5}) = -3\lambda\) Next, divide both sides by -3 to isolate \(\lambda\): \(\lambda = -\frac{1}{3} ln(\frac{4}{5})\)
03

Relate decay constant 位 to half-life T

The relationship between the decay constant \(\lambda\) and half-life T is given by: \(\lambda = \frac{ln(2)}{T}\) Now substitute the value of 位 obtained in step 2 into this equation: \(-\frac{1}{3} ln(\frac{4}{5}) = \frac{ln(2)}{T}\)
04

Solve for the half-life T

To find T, we first multiply both sides by 3: \(-ln(\frac{4}{5}) = 3\frac{ln(2)}{T}\) Next, multiply both sides by T: \(-T\cdot ln(\frac{4}{5}) = 3\cdot ln(2)\) Finally, divide both sides by -ln(4/5) to solve for T: \(T = \frac{3\cdot ln(2)}{-ln(\frac{4}{5})}\)
05

Calculate the half-life T

Now we can plug in the values to calculate the half-life: \(T = \frac{3\cdot ln(2)}{-ln(\frac{4}{5})} 鈮� 7.42\) So the half-life of this radioactive isotope is approximately 7.42 years.

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