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Iodine-131 has a half-life of 8.0 days. How many days will it take for 174 g of \(^{131}\) I to decay to 83 g of \(^{131}\) I?

Short Answer

Expert verified
It will take approximately 8.0 days for 174 grams of Iodine-131 to decay to 83 grams.

Step by step solution

01

Setup the half-life decay formula

Write down the half-life decay formula: Final amount = Initial amount * (1/2)^(Time elapsed / Half-life) Insert given values: 83 g = 174 g * (1/2)^(Time elapsed / 8.0 days)
02

Solve for Time elapsed

First, divide both sides of the equation by 174 g to isolate the exponential term: (83 g) / (174 g) = (1/2)^(Time elapsed / 8.0 days) Now take the logarithm base 2 of both sides: log2( (83 g) / (174 g) ) = log2( (1/2)^(Time elapsed / 8.0 days) ) Using the property of logarithms, "log(a^x) = x * log(a)", the equation simplifies to: log2( (83 g) / (174 g) ) = (Time elapsed / 8.0 days) * log2(1/2) Divide both sides by log2(1/2), which is -1: Time elapsed / 8.0 days = -log2( (83 g) / (174 g) ) Finally, multiply both sides by the half-life (8.0 days) to find the Time elapsed: Time elapsed = (-log2( (83 g) / (174 g) )) * 8.0 days
03

Calculate the result

Plug the values into a calculator and compute the Time elapsed: Time elapsed = (-log2( 83 / 174 )) * 8.0 ≈ 8.0 days It will take approximately 8.0 days for 174 grams of Iodine-131 to decay to 83 grams.

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Key Concepts

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

Iodine-131 Decay
Understanding the decay of substances like Iodine-131 is crucial in numerous fields such as medicine, where it's used for treating thyroid cancer and in medical diagnostics. Over time, Iodine-131 decays into a more stable element.

Iodine-131 is a radioactive isotope that decays following a predictable pattern, characterized by its half-life. The half-life is the amount of time it takes for half of a given quantity of the isotope to decay. In the case of Iodine-131, this period is 8 days. This means that every 8 days, half of the substance will have transformed into a different isotope or element through the process of radioactive decay.
Exponential Decay
The concept of exponential decay is widely observed in natural processes, including the decay of radioactive substances. It describes a situation where a quantity decreases at a rate proportional to its current value.

In the context of Iodine-131, exponential decay means the amount remaining will decrease by half over each half-life period. The mathematical model of exponential decay is expressed through the formula \( N(t) = N_0 \times (1/2)^{\frac{t}{t_{1/2}}} \), where \( N(t) \) is the quantity at time \( t \), \( N_0 \) is the initial quantity, and \( t_{1/2} \) is the half-life. This formula shows how the quantity of a substance changes over time using an exponent that involves the half-life.
Radioactive Decay
Radioactive decay is a spontaneous process by which an unstable atomic nucleus loses energy by emitting radiation. In the decay process, isotopes transform into more stable forms, either by shedding particles or through processes like gamma decay, where energy is released without a change in particle count.

This release of energy and particles can be hazardous, which is why understanding and calculating the decay of radioactive substances like Iodine-131 is essential for safety in nuclear medicine and other industries using radioactive materials. The exponential decay model is key in predicting when the material becomes safe or is at an acceptable level of radioactivity for medical and industrial applications.
Logarithms in Chemistry
Logarithms are mathematical tools that help in unraveling exponential relationships, which are prevalent in chemistry, especially concerning reaction rates and radioactive decay.

In the context of half-life calculations, logarithms allow us to solve for the time elapsed during decay. The half-life equation can be manipulated using logarithms to isolate \( t \), the time variable. This is vital in real-world applications, such as calculating the dosage and timing for radioisotope treatments in medicine, or estimating the safe periods for workers to handle materials in nuclear facilities.

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Most popular questions from this chapter

Radioactive cobalt-60 is used to study defects in vitamin \(\mathbf{B}_{12}\) absorption because cobalt is the metallic atom at the center of the vitamin \(\mathrm{B}_{12}\) molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron-58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: \(^{60} \mathrm{Co}=\) \(\left.59.9338 \mathrm{u} ;^{1} \mathrm{H}=1.00782 \mathrm{u}\right) ?\) What is the de Broglie wave-length of the emitted particle if it has a velocity equal to \(0.90 c\) where \(c\) is the speed of light?

In addition to the process described in the text, a second process called the carbon-nitrogen cycle occurs in the sun:a. What is the catalyst in this process? b. What nucleons are intermediates? c. How much energy is released per mole of hydrogen nuclei in the overall reaction? (The atomic masses of \(i \mathrm{H}\) and 4 He are 1.00782 u and 4.00260 u, respectively.)

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \(\frac{3}{1} \mathrm{H}(\beta)\) b. \(\frac{8}{3} L i(\beta \text { followed by } \alpha)\) c. \(^{7}_{4}\) Be (electron capture)d. \(^{8}_{5} B\) (positron)

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July \(16,1945 .\) What percentage of the strontium- \(90(t_{1 / 2}=28.9\) years) originally produced . by that explosion still remains as of July \(16,2015 ?\)

A chemist studied the reaction mechanism for the reaction $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$$ by reacting \(\mathrm{N}^{16} \mathrm{O}\) with \(^{18} \mathrm{O}_{2}\). If the reaction mechanism is $$\begin{aligned} \mathrm{NO}+\mathrm{O}_{2} & \rightleftharpoons \mathrm{NO}_{3}(\text { fast equilibrium }) \\ \mathrm{NO}_{3}+\mathrm{NO} & \longrightarrow 2 \mathrm{NO}_{2}(\text { slow }) \end{aligned}$$ what distribution of \(^{18} \mathrm{O}\) would you expect in the \(\mathrm{NO}_{2} ? \)Assume that \(\mathrm{N}\) is the central atom in \(\mathrm{NO}_{3},\) assume only \(\mathrm{N}^{16} \mathrm{O}^{18} \mathrm{O}_{2}\) forms, and assume stoichiometric amounts of reactants are combined.

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