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

The half-life of a radioactive isotope is \(1.5\) hours. The mass of it that remains undecayed after 6 hours is (the initial mass of the isotope is \(64 \mathrm{~g}\) ) (a) \(32 \mathrm{~g}\) (b) \(16 \mathrm{~g}\) (c) \(8 \mathrm{~g}\) (d) \(4 \mathrm{~g}\)

Short Answer

Expert verified
Option (d) 4 g.

Step by step solution

01

Understanding Half-Life

The half-life of a radioactive isotope is the time it takes for half of the material to decay. For this isotope, the half-life is given as 1.5 hours.
02

Total Time Calculation

We are given a duration of 6 hours, and we need to find out how many half-lives fit into this total time. Divide the total time (6 hours) by the half-life (1.5 hours). \[\text{Number of half-lives} = \frac{6}{1.5} = 4\]
03

Calculate Remaining Mass

The number of half-lives determines how many times the mass is halved. Start with the initial mass of 64 grams and halve it four times because of the 4 half-lives.After 1st half-life: \[\frac{64}{2} = 32 \text{g}\]After 2nd half-life: \[\frac{32}{2} = 16 \text{g}\]After 3rd half-life: \[\frac{16}{2} = 8 \text{g}\]After 4th half-life: \[\frac{8}{2} = 4 \text{g}\]
04

Select the Correct Option

From the calculations, the mass remaining after 6 hours is 4 grams. Comparing with the given options, the correct answer is (d) \(4 \text{ g}\).

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.

Half-Life
The concept of half-life is crucial in understanding radioactive decay. When a radioactive isotope decays, it transforms into a different, more stable state by emitting particles. The half-life of an isotope is the time it takes for half of the original substance to decay. This can vary widely across different isotopes—some have half-lives of mere seconds while others stretch across thousands of years.
To visualize this, imagine having a pile of 64 grams of a radioactive isotope. With a half-life of 1.5 hours, it means that in every 1.5-hour interval, the remaining mass of the isotope will be reduced by half. In the given exercise, after 6 hours, the initial mass will have gone through four half-lives, leaving only 4 grams of the isotope remaining from the original 64 grams.
Isotope
An isotope is a variation of an element that has the same number of protons but a different number of neutrons. This change in neutron number can lead to various properties, most notably radioactivity in some cases.
For instance, carbon-12 and carbon-14 are both isotopes of carbon. While carbon-12 is stable, carbon-14 is radioactive and undergoes decay over time, making it useful in techniques such as radiocarbon dating. In this scenario, the term 'radioactive isotope' refers to these particular forms of an element that are not stable and will change over time by emitting radiation.
Radioactive isotopes have wide-ranging applications, including medical treatments, archaeological dating, and energy production in nuclear reactors.
Mass Calculation
Mass calculation during radioactive decay is a step-by-step process that helps us understand how much of a material remains after a certain time. Knowing the half-life allows you to determine how many times the mass is halved over a given duration.
The exercise provided used a straightforward mass-halving method by dividing the remaining mass by two after each half-life. Starting with 64 grams and reducing it every 1.5 hours through four half-lives led to a final undecayed mass of 4 grams.
This calculation is straightforward because each step reduces the mass by a consistent and predictable amount—half of the previous mass—demonstrating the exponential nature of decay in radioactive isotopes.
Radioactivity Concepts
Radioactivity involves the spontaneous emission of particles or energy from an unstable atomic nucleus. This natural occurrence is the underlying principle of radioactive decay, observed when isotopes transform into more stable forms.
Key types of decay include alpha, beta, and gamma decay, each involving different particles and energy types. Understanding radioactivity concepts is essential as it has numerous practical applications, including:
  • Medical Imaging and Treatment: Using radioactive isotopes to detect and treat conditions like cancer.
  • Radiometric Dating: Helping scientists determine the age of ancient objects and geological formations.
  • Energy Production: Harnessing nuclear reactions from isotopes in power plants to generate electricity.
Grasping these concepts is vital for appreciating the processes and technologies that utilize radioactivity to benefit society.

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

A nuclide of an alkaline earth metal undergoes radioactive decay by emission of the \(\alpha\) particle in succession. The group of the periodic table to which the resulting daughter element would belong is (a) Gp 14 (b) Gp 6 (c) Gp 16 (d) Gp 4

A sample of \(\mathrm{U}^{238}\left(\mathrm{t}_{1 / 2}=4.5 \times 10^{9} \mathrm{yrs}\right)\) ore is found con- taining \(23.8 \mathrm{~g} \mathrm{U}^{238}\) and \(20.6 \mathrm{~g}\) of \(\mathrm{Pb}^{206} .\) Calculate the age of the ore. (a) \(4.9 \times 10^{9}\) year (b) \(9.0 \times 10^{11}\) year (c) \(9.4 \times 10^{9}\) year (d) \(4.5 \times 10^{9}\) year

A photon of hard \(\gamma\) radiation knocks a proton out of \({ }_{12} \mathrm{Mg}^{44}\) nucleus to form (a) the isotope of parent nucleus (b) the isobar of parent nucleus (c) the nuclide of \({ }_{11} \mathrm{Na}^{23}\) (d) the isobar of \(_{11} \mathrm{Na}^{23}\)

At radioactive equilibrium, the ratio between the atoms of two radioactive elements \(\mathrm{X}\) and \(\mathrm{Y}\) was found to be \(3.1 \times 10^{9}: 1\) respectively. If \(\mathrm{T}_{50}\) of the element \(\mathrm{X}\) is \(2 \times 10^{10}\) years, then \(\mathrm{T}_{50}\) of the element \(\mathrm{Y}\) is (a) \(6.45\) years (b) \(3.1 \times 10^{6}\) years (c) \(6.2 \times 10^{7}\) years (d) \(21 \times 10^{8}\) years

The radio nucliede \({ }_{90} \mathrm{Th}^{234}\) undergoes two successive \(\beta\) decays followed by one \(\alpha\) decay. The atomic number and the mass number respectively of the resulting radio nucliede will be (a) 92 and 234 (b) 94 and 230 (c) 90 and 230 (d) 92 and 230

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