Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Two isotopes 'P' and 'Q' of atomic masses 10 and 20, respectively, are mixed in equal amount, by mass. After 20 day, their mass ratio is found to be \(1: 4 .\) Isotope 'P' has a half-life of 10 days. The half-life of isotope 'Q' is (a) zero (b) 5 day (c) 20 day (d) infinite

Short Answer

Expert verified
The half-life of isotope 'Q' is infinite.

Step by step solution

01

Understanding the Concept of Half-Life

The half-life of a radioactive substance is the time it takes for its amount to reduce to half its original amount. Since isotope 'P' has a half-life of 10 days, it means that after 10 days, only half of it remains, and after another 10 days (a total of 20 days), only a quarter of it will remain.
02

Calculating the Remaining Mass of Isotope 'P' after 20 Days

After 20 days, which is two half-lives for isotope 'P', the amount of isotope 'P' left would be one quarter of its original amount. Therefore, if the original mass of isotope 'P' is m, the remaining mass after 20 days would be \(\frac{m}{4}\).
03

Analyzing the Mass Ratio of P and Q after 20 Days

According to the problem, the mass ratio of isotope 'P' to isotope 'Q' after 20 days is 1:4. This means that for every 1 part of isotope 'P', there are 4 parts of isotope 'Q'. Let's say the original mass of isotope 'Q' is also m. If the half-life of 'Q' were finite and less than 20 days, then 'Q' would have decayed by some factor, and the ratio would have been different.
04

Determining the Half-Life of Isotope 'Q'

The only way the mass ratio can be 1:4 after 20 days is if isotope 'Q' has not decayed at all, which implies that its half-life must be infinite. Therefore, option (d) infinite is the correct answer.

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.

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

Key Concepts

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

Physical Chemistry and Radioactive Half-Life
Physical chemistry involves the study of how matter behaves on a molecular and atomic level, and it explains how chemical reactions occur. In the context of radioactive half-life problems, physical chemistry looks at the kinetics of nuclear reactions and the rates at which unstable isotopes decay.

Understanding the half-life concept is essential in physical chemistry, especially when dealing with radioactive substances. The half-life is the time required for half of a sample of a radioactive isotope to decay. This decay process follows first-order kinetics, which means the rate of decay is proportional to the number of undecayed nuclei present. As a result, in each succeeding half-life period, exactly half of the remaining radioactive atoms will decay, no matter how many atoms you started with.

Knowing this property allows chemists and physicists to predict the behavior of radioactive materials over time, calculate the age of archaeological artifacts, or even estimate the dosage of radioactive medicine in treatments such as radiation therapy.
Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. A material containing such unstable nuclei is considered radioactive. There are several types of radioactive decay, including alpha decay, beta decay, and gamma decay, each involving the emission of different particles or photons.

The rate of decay is exponential, and this is quantitatively described by what we call a 'decay constant'. The larger the decay constant, the more quickly the isotope will decay. The concept of half-life is inversely related to the decay constant; a short half-life indicates a fast decay rate. An infinite half-life would suggest that the isotope is effectively stable and does not undergo radioactive decay over observable timeframes. This is a key concept when considering the stability of isotopic samples and their potential uses or hazards in various applications.
Isotopic Mass Ratio
In physical and chemical sciences, the isotopic mass ratio is a comparison of the amount of different isotopes of a particular element present in a sample. In the context of a half-life problem, this mass ratio changes over time as different isotopes decay at different rates.

By knowing the original isotopic mass ratio and the half-life of each isotope, we can calculate how this ratio will change over time. This becomes particularly useful in the fields of geology, archaeology, and environmental science, where the isotopic mass ratio can tell us about the age of a sample or help us trace the origin of certain materials. The problem presented on radioactive half-life demonstrates how the isotopic mass ratio shifts due to radioactive decay—a fundamental concept for students learning about radiochemistry and its applications. By understanding this concept, one can also solve complex problems related to nuclear medicine, carbon dating, and nuclear power.

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

Bismuth is the end product of radioactive disintegration series known as (a) \(4 n\) (b) \(4 n+1\) (c) \(4 n+2\) (d) \(4 n+3\)

If \(8 \mathrm{~g}\) of a radioactive isotope has a halflife of \(10 \mathrm{~h}\). The half-life of \(2 \mathrm{~g}\) of the same substance is (a) \(2.5 \mathrm{~h}\) (b) \(5 \mathrm{~h}\) (c) \(10 \mathrm{~h}\) (d) \(40 \mathrm{~h}\)

In nature a decay chain starts with \(\mathrm{Th}^{232}\) and finally terminates at \(\mathrm{Pb}^{208}\). A thorium ore sample was found to contain \(6.72 \times 10^{-5} \mathrm{ml}\) of \(\mathrm{He}\) (at \(273 \mathrm{~K}\) and \(\left.1 \mathrm{~atm}\right)\) and \(4.64 \times 10^{-7} \mathrm{~g}\) of \(\mathrm{Th}^{232}\). Find the age of the sample assuming that source of He to be only due to decay of \(\mathrm{Th}^{232}\). Also assume complete retention of He within the ore. \(\left(t_{1 / 2}\right.\) of \(\mathrm{Th}^{232}=1.38 \times 10^{10}\) years, \(\log 2=0.3)\) (a) \(2.3 \times 10^{10}\) years (b) \(2.3 \times 10^{9}\) years (c) \(4.6 \times 10^{9}\) years (d) \(9.2 \times 10^{9}\) years

A free neutron decays to a proton but a free proton does not decay to a neutron. This is because (a) neutron is a composite particle made of a proton and an electron whereas proton is a fundamental particle (b) neutron is a uncharged particle whereas proton is a charged particle (c) neutron has larger rest mass than a proton (d) weak forces can operate in a neutron but not in a proton

Which element has a relatively more stable nucleus? (a) U (b) \(\mathrm{H}\) (c) Fe (d) Ra

See all solutions

Recommended explanations on Chemistry Textbooks

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