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Chapter 29: Problem 34

The half-life of I-131 is 8.0 days. A sample containing I- 131 has an activity of \(6.4 \times 10^{8}\) Bq. How many days later will the sample have an activity of \(2.5 \times 10^{6} \mathrm{Bq} ?\) (Wills tutorial: radioactive decay)

Short Answer

Expert verified
Answer: It takes approximately 29.6 days for the I-131 sample's activity to reduce from \(6.4 \times 10^{8} \mathrm{Bq}\) to \(2.5 \times 10^{6} \mathrm{Bq}\).

Step by step solution

01

Calculate the decay constant \(\lambda\)

Using the half-life formula: \(\lambda = \frac{ln(2)}{t_{1/2}}\), we can find the decay constant. \(t_{1/2} = 8.0 \, \text{days}\) (given) \(\lambda = \frac{ln(2)}{8.0}\)
02

Set up the radioactive decay equation

Now that we have the decay constant, λ, we can use the radioactive decay formula to find the time t. \(A = A_0 e^{-\lambda t}\) We are given: - \(A = 2.5 \times 10^{6} \, \text{Bq}\) - \(A_0 = 6.4 \times 10^{8} \, \text{Bq}\) We will now plug these values into the equation: \(2.5 \times 10^{6} = 6.4 \times 10^{8} e^{-\lambda t}\)
03

Solve for t

To solve for \(t\), we first isolate \(e^{-\lambda t}\) by dividing both sides by \(6.4 \times 10^{8}\): \(e^{-\lambda t} = \frac{2.5 \times 10^{6}}{6.4 \times 10^{8}}\) Now, we will take the natural logarithm of both sides to isolate the \(-\lambda t\) term: \(-\lambda t = ln\left(\frac{2.5 \times 10^{6}}{6.4 \times 10^{8}}\right)\) Finally, we divide by \(-\lambda\) to solve for \(t\). Since \(\lambda = \frac{ln(2)}{8}\), we have: \(t = \frac{ln\left(\frac{2.5 \times 10^{6}}{6.4 \times 10^{8}}\right)}{-\frac{ln(2)}{8}}\)
04

Calculate the time

Now we will calculate the value of \(t\) using the values we derived: \(t = \frac{ln\left(\frac{2.5 \times 10^{6}}{6.4 \times 10^{8}}\right)}{-\frac{ln(2)}{8}} = \boxed{29.6} \, \text{days}\) (approximately) So, it would take around 29.6 days for the I-131 sample's activity to reduce from \(6.4 \times 10^{8} \mathrm{Bq}\) to \(2.5 \times 10^{6} \mathrm{Bq}\).

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Most popular questions from this chapter

What is the activity in becquerels of \(1.0 \mathrm{kg}\) of $^{238} \mathrm{U} ?$
A radioactive sample has equal numbers of \(^{15} \mathrm{O}\) and $^{19} \mathrm{O}$ nuclei. Use the half-lives found in Appendix B to determine how long it will take before there are twice as many \(^{15} \mathrm{O}\) nuclei as \(^{19} \mathrm{O} .\) What percent of the \(^{19} \mathrm{O}\) nuclei have decayed during this time?
An \(\alpha\) particle with a kinetic energy of \(1.0 \mathrm{MeV}\) is headed straight toward a gold nucleus. (a) Find the distance of closest approach between the centers of the \(\alpha\) particle and gold nucleus. (Assume the gold nucleus remains stationary. since its mass is much larger than that of the \(\alpha\) particle, this assumption is a fairly good approximation.) (b) Will the two get close enough to "touch"? (c) What is the minimum initial kinetic energy of an \(\alpha\) particle that will make contact with the gold nucleus?
The radioactive decay of \(^{238} \mathrm{U}\) produces \(\alpha\) particles with a kinetic energy of \(4.17 \mathrm{MeV} .\) (a) At what speed do these \(\alpha\) particles move? (b) Put yourself in the place of Rutherford and Geiger. You know that \(\alpha\) particles are positively charged (from the way they are deflected in a magnetic field). You want to measure the speed of the \(\alpha\) particles using a velocity selector. If your magnet produces a magnetic field of \(0.30 \mathrm{T},\) what strength electric field would allow the \(\alpha\) particles to pass through undeflected? (c) Now that you know the speed of the \(\alpha\) particles, you measure the radius of their trajectory in the same magnetic field (without the electric field) to determine their charge-to-mass ratio. Using the charge and mass of the \(\alpha\) particle, what would the radius be in a \(0.30-\mathrm{T}\) field? (d) Why can you determine only the charge-to-mass ratio \((q / m)\) by this experiment, but not the individual values of \(q\) and \(m ?\)
Approximately what is the total energy of the neutrino emitted when ${ }_{11}^{22} \mathrm{Na}$ decays by electron capture?
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