Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 71

Define third-life in a similar way to half-life, and determine the "third- life" for a nuclide that has a half-life of 31.4 years.

Short Answer

Expert verified
The third-life, similar to half-life, is the time required for a quantity to reduce to one-third of its initial value. To calculate the third-life for a nuclide with a half-life of 31.4 years, first determine its decay constant, \(λ\), using the half-life equation: \(λ = \frac{ln(2)}{t_{1/2}} ≈ 0.0221\text{ year}^{-1}\). Then, calculate the third-life using the equation \(t_{1/3} = \frac{ln(3)}{λ} ≈ 49.85\text{ years}\).

Step by step solution

01

Define third-life

Third-life can be defined as the time required for a quantity (in this case, the number of radioactive atoms in a sample of a nuclide) to reduce to one-third its initial value.
02

Understand the concept of half-life

Half-life is the time required for a quantity to reduce to half its initial value. In nuclear decay, it is often used to describe the time required for half of the radioactive atoms in a sample to decay. We are given the half-life of a nuclide as 31.4 years.
03

Calculate the decay constant

To find the third-life, we'll first need to calculate the decay constant of the nuclide. The decay constant is represented as λ (lambda). It can be calculated using the half-life equation: \[λ = \frac{ln(2)}{t_{1/2}}\] where \(t_{1/2}\) is the half-life \(ln(2)\) is the natural logarithm of 2 Now, plug in the given half-life value: \[λ = \frac{ln(2)}{31.4}\] \[λ ≈ 0.0221\text{ year}^{-1}\]
04

Calculate the third-life

Next, we'll use the decay constant to find the third-life. We can define third-life (\(t_{1/3}\)) using the decay constant in a similar equation as the half-life equation: \[t_{1/3} = \frac{ln(3)}{λ}\] where \(ln(3)\) is the natural logarithm of 3 Now, plug in the previously calculated decay constant: \[t_{1/3} = \frac{ln(3)}{0.0221}\] \[t_{1/3} ≈ 49.85\text{ years}\] So, the third-life for the given nuclide is approximately 49.85 years.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which type of radioactive decay has the net effect of changing a neutron into a proton? Which type of decay has the net effect of turning a proton into a neutron?

Describe the relative penetrating powers of alpha, beta, and gamma radiation.

Scientists have estimated that the earth's crust was formed 4.3 billion years ago. The radioactive nuclide \(176 \mathrm{Lu},\) which decays to 176 \(\mathrm{Hf}\) , was used to estimate this age. The half-life of 176 \(\mathrm{Lu}\) is 37 billion years. How are ratios of \(^{176} \mathrm{Lu}\) to 176 \(\mathrm{Hf}\) utilized to date very old rocks?

What are transuranium elements and how are they synthesized?

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 \(^{24} \mathrm{Am}\) 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 \(^{241} \mathrm{Am}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free