Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 27

The radioactive decay of T1-206 to \(\mathrm{Pb}-206\) has a half-life of 4.20 min. Starting with \(5.00 \times 10^{22}\) atoms of T1-206, calculate the number of such atoms left after \(42.0 \mathrm{~min} .\)

Short Answer

Expert verified
After 42.0 minutes, there will be approximately \(4.88 \times 10^{19}\) atoms of Tl-206 remaining.

Step by step solution

01

Identify given figures

The given half-life of Tl-206 is 4.20 minutes and we start with \(5.00 \times 10^{22}\) atoms. We are asked to find the remaining quantity after 42.0 minutes.
02

Calculate the number of half-lives

Given that each half-life lasts for 4.20 minutes, to find out how many half-lives have passed in 42.0 minutes, divide 42.0 min by 4.20 min/half-life to obtain 10 half-lives.
03

Apply the formula for decay using half-life

The remaining quantity (Q) of a substance after n half-lives is given by \(Q = Q_0 \times (\frac{1}{2})^n\) where \(Q_0\) is the initial quantity and n is the number of half-lives. In this case, \(Q_0 = 5.00 \times 10^{22}\) atoms and n = 10 half-lives. So, \(Q = 5.00 \times 10^{22} \times (\frac{1}{2})^{10}\).

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

After the Chernobyl accident, people living close to the nuclear reactor site were urged to take large amounts of potassium iodide as a safety precaution. What is the chemical basis for this action?

The following equations are for nuclear reactions that are known to occur in the explosion of an atomic bomb. Identify X. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{140} \mathrm{Ba}+3{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (b) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{55}^{144} \mathrm{Cs}+{ }_{37}^{90} \mathrm{Rb}+2 \mathrm{X}\) (c) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{35}^{87} \mathrm{Br}+3{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (d) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{62}^{160} \mathrm{Sm}+{ }_{30}^{72} \mathrm{Zn}+4 \mathrm{X}\).

For each pair of isotopes listed, predict which one is less stable: (a) \({ }_{3}^{6} \mathrm{Li}\) or \({ }_{3}^{9} \mathrm{Li}\), (b) \({ }_{11}^{23} \mathrm{Na}\) or \({ }_{11}^{25} \mathrm{Na}\) (c) \({ }_{20}^{48} \mathrm{Ca}\) or \({ }_{21}^{48} \mathrm{Sc}\).

How do nuclear reactions differ from ordinary chemical reactions?

A freshly isolated sample of \({ }^{90} \mathrm{Y}\) was found to have an activity of \(9.8 \times 10^{5}\) disintegrations per minute at 1: 00 P.M. on December 3,2003 . At 2: 15 P.M. on December \(17,2003,\) its activity was redetermined and found to be \(2.6 \times 10^{4}\) disintegrations per minute. Calculate the half-life of \({ }^{90} \mathrm{Y}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

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