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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

Fresh rainwater or surface water contains enough tritium ( \(^{3}_{1} \mathrm{H}\) ) to show 5.5 decay events per minute per \(100 .\) g water. Tritium has a half-life of 12.3 years. You are asked to check a vintage wine that is claimed to have been produced in \(1946 .\) How many decay events per minute should you expect to observe in \(100 .\) g of that wine?

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \(^{241} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a 5.00 -g sample of \(^{241} \mathrm{Am} ?\)

Given the following information: Mass of proton \(=1.00728 \mathrm{u}\) Mass of neutron \(=1.00866 \mathrm{u}\) Mass of electron \(=5.486 \times 10^{-4} \mathrm{u}\) Speed of light \(=2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\) Calculate the nuclear binding energy of \(\frac{24}{12} \mathrm{Mg},\) which has an atomic mass of 23.9850 u.

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn T. Seaborg, discoverer of the transuranium elements.a. \(^{263} \mathrm{Sg}\) was produced by the bombardment of \(^{249} \mathrm{Cf}\) with a beam of \(^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \(^{263}\) Sg decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg} ?\)

Estimate the temperature needed to achieve the fusion of deuterium to make an \(\alpha\) particle. The energy required can be estimated from Coulomb's law [use the form \(E=9.0 \times 10^{9}\) \(\left(Q_{1} Q_{2} / r\right),\) using \(Q=1.6 \times 10^{-19} \mathrm{C}\) for a proton, and \(r=2 \times\) \(10^{-15} \mathrm{m}\) for the helium nucleus; the unit for the proportionality constant in Coloumb's law is \(\left.\mathbf{J} \cdot \mathbf{m} / \mathbf{C}^{2}\right]\).

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