It takes 4 \(\mathrm{h} 39\) min for a 2.00 - mg sample of radium-230 to decay to 0.25 \(\mathrm{mg.}\) What is the half-life of radium-230?

Short Answer

Expert verified
The half-life of radium-230 is 93 minutes.

Step by step solution

01

Convert the given time to minutes

We have 4 hours and 39 minutes. Convert this time to minutes by multiplying the hours by 60 and adding the minutes. Time = (4 hours × 60 minutes/hour) + 39 minutes = 240 + 39 = 279 minutes
02

Write down the decay formula

The decay formula is given by the equation: \[N(t) = N_0 \times 2^{-\frac{t}{T}}\] where - \(N(t)\) is the final mass of the substance at time \(t\), - \(N_0\) is the initial mass of the substance, - \(t\) is the time of decay, - \(T\) is the half-life period.
03

Substitute the given values into the formula

We have \(N(t) = 0.25\,\mathrm{mg}\), \(N_0 = 2.00\,\mathrm{mg}\), and \(t = 279\,\mathrm{min}\). Substitute these values into the decay formula: \[\begin{align*}0.25 &= 2 \times 2^{-\frac{279}{T}}\end{align*}\]
04

Solve for the half-life period \(T\)

Isolate \(2^{-\frac{279}{T}}\) by dividing both sides by 2: \[\begin{align*}2^{-\frac{279}{T}} &= \frac{1}{8}\end{align*}\] Now we can take the logarithm to base 2 on both sides to remove the exponent: \[^{-\frac{279}{T}} = \log_2\left(\frac{1}{8}\right)\] Therefore, \(\,\frac{279}{T}\) will be equal to the result of \(\log_2\left(\frac{1}{8}\right)\): \[\frac{279}{T} = -3\] Now, solve for \(T\): \[T = \frac{279}{3} = 93\]
05

Write the final answer

The half-life of radium-230 is 93 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Decay
Radioactive decay is a fundamental concept in nuclear chemistry, involving the spontaneous transformation of an unstable atomic nucleus into a more stable one. This process emits radiation in the form of particles or electromagnetic waves and occurs naturally in a variety of elements found on Earth.

Substances that undergo radioactive decay are called radionuclides, and they possess what's known as a half-life, which is the time required for half of the radioactive atoms present in a sample to decay. Radioactive decay is a crucial process in fields such as medicine, where it's used for diagnostic imaging and cancer treatment, as well as in carbon dating, where it helps determine the age of archeological finds.
Decay Formula
Understanding the decay formula is essential in solving problems involving radioactive decay. This formula, represented as
\[N(t) = N_0 \times 2^{-\frac{t}{T}}\],
describes the relationship between the remaining quantity of a substance
(\(N(t)\)) after time
(\(t\)), its initial quantity
(\(N_0\)), and the half-life period
(\(T\)). It’s vital to note that the exponent –\(\frac{t}{T}\) indicates how many half-lives have elapsed. The role of this exponential factor is to calculate how much of the initial sample remains after a given period.
Logarithms in Decay
Logarithms play a crucial role in simplifying the process of solving for the half-life period in decay problems. By applying logarithms, we can transform the exponential decay equation into a linear form, making the half-life easier to calculate.

In our example, taking the logarithm to base 2 of both sides enables us to isolate the half-life
(\(T\)), bypassing the complexity of dealing with exponents directly. This is done because logarithms have the unique property of turning multiplication and powers into addition and multiplication, respectively, thus untangling the variables entwined by the exponential function and allowing for straightforward computation.
Nuclear Chemistry
Nuclear chemistry encompasses the study of radioactive substances and the chemical changes that occur due to nuclear reactions. It involves understanding the processes that alter the composition of atomic nuclei and has vast applications, from generating electricity in nuclear power plants to developing treatments that target cancer cells.

In educational terms, diving into this branch of chemistry equips students with the knowledge of how elements can transform and the effects of such transformations, both scientifically and environmentally. Moreover, it underscores the importance of safety and regulations when dealing with materials that emit radiation, reinforcing the need for responsible practices in this powerful field of science.

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Most popular questions from this chapter

Write balanced equations for each of the following nuclear reactions: \((\mathbf{a}) _{92}^{238} \mathrm{U}(\mathrm{n}, \gamma)_{92}^{239} \mathrm{U},\) \((\mathbf{b})_{8}^{16} \mathrm{O}(\mathrm{p}, \alpha)_{7}^{13} \mathrm{N},\) \((\mathbf{c})_{8}^{18} \mathrm{O}\left(\mathrm{n}, \beta^{-}\right)_{9}^{19} \mathrm{F}.\)

Why is it important that radioisotopes used as diagnostic tools in nuclear medicine produce gamma radiation when they decay? Why are alpha emitters not used as diagnostic tools?

(a) Which of the following are required characteristics of an isotope to be used as a fuel in a nuclear power reactor? (i) It must emit gamma radiation. (ii) On decay, it must release two or more neutrons. (iii) It must have a half-life less than one hour. (iv) It must undergo fission upon the absorption of a neutron. (b) What is the most common fissionable isotope in a commercial nuclear power reactor?

A 25.0 -mL sample of 0.050\(M\) barium nitrate solution was mixed with 25.0 mL of 0.050\(M\) sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was \(1.22 \times 10^{6} \mathrm{Bq} / \mathrm{mL}\) . After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 250 \(\mathrm{Bq} / \mathrm{mL}\) . (a) Write a balanced chemical equation for the reaction that occurred. (b) Calculate the \(K_{s p}\) for the precipitate under the conditions of the experiment.

The steps below show three of the steps in the radioactive decay chain for \(_{90}^{232}\) Th. The half-life of each isotope is shown below the symbol of the isotope. (a) Identify the type of radioactive decay for each of the steps (i), (ii), and (iii). (b) Which of the isotopes shown has the highest activity? (c) Which of the isotopes shown has the lowest activity? (d) The next step in the decay chain is an alpha emission. What is the next isotope in the chain? [Sections 21.2 and 21.4]

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