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

Why are nuclear transmutations involving neutrons generally easier to accomplish than those involving protons or alpha particles?

Short Answer

Expert verified
Nuclear transmutations involving neutrons are generally easier to accomplish than those involving protons or alpha particles because neutrons have no electrical charge, resulting in no electrostatic repulsion and lower energy requirements to penetrate the nucleus. Additionally, neutrons can contribute to the stability of the resultant nucleus, making these transmutations more likely to result in stable nuclei.

Step by step solution

01

Consider electrostatic repulsion

To induce nuclear transmutations, it is necessary to overcome the electrostatic repulsion force between the charged particles. Neutrons carry no electrical charge, and therefore, they do not experience any electrostatic repulsion when approaching the nucleus of an atom. In contrast, protons and alpha particles (which consist of two protons and two neutrons) carry a positive charge leading to a repulsive force with the target nucleus. This makes it more difficult for protons or alpha particles to approach the target nucleus and take part in nuclear transmutations.
02

Analyze the energy requirements for transmutations

When it comes to nuclear reactions, the energy barrier plays a crucial role. As neutrons have no electric charge, they can easily penetrate the nucleus without need for a high amount of energy, thus making the nuclear transmutations more likely to occur. On the other hand, charged particles like protons and alpha particles need to overcome the energy barrier created by the repulsive electrostatic forces. As a result, transmutations involving protons or alpha particles generally require higher energy to occur.
03

Consider the stability of the resultant nucleus

Nuclear stability depends on the balance between the number of protons and neutrons in a nucleus. Neutrons can help stabilize the nucleus by compensating for the repulsive electrostatic force between protons. Additionally, neutrons can bind to protons to form more stable nuclei, whereas the addition of protons or alpha particles may lead to an unstable nucleus. So, transmutations involving neutrons are more likely to result in stable nuclei than those involving protons or alpha particles. In conclusion, nuclear transmutations involving neutrons are generally easier to accomplish compared to those involving protons or alpha particles due to the absence of electrostatic repulsion, lower energy requirements, and the higher likelihood of producing a stable nucleus.

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!

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

Methyl acetate \(\left(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right)\) is formed by the reaction of acetic acid with methyl alcohol. If the methyl alcohol is labeled with oxygen-18, the oxygen-18 ends up in the methyl acetate: Do the \(\mathrm{C}-\mathrm{OH}\) bond of the acid and the \(\mathrm{O}-\mathrm{H}\) bond of the alcohol break in the reaction, or do the \(\mathrm{O}-\mathrm{H}\) bond of the acid and the \(\mathrm{C}-\mathrm{OH}\) bond of the alcohol break? Explain.

A radioactive decay series that begins with \({ }_{9}^{23} \mathrm{Th}\) ends with formation of the stable nuclide \({ }_{82}^{208} \mathrm{~Pb}\). How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

Americium-241 is an alpha emitter used in smoke detectors. The alpha radiation ionizes molecules in an air-filled gap between two electrodes in the smoke detector, leading to current. When smoke is present, the ionized molecules bind to smoke particles and the current decreases; when the current is reduced sufficiently, an alarm sounds. (a) Write the nuclear equation corresponding to the alpha decay of americium-241. (b) Why is an alpha emitter a better choice than a gamma emitter for a smoke detector? (c) In a commercial smoke detector, only \(0.2\) micrograms of americium are present. Calculate the energy that is equivalent to the mass loss of this amount of americium due to alpha radiation. The atomic mass of americium- 241 is \(241.056829\) amu. (d) The half-life of americium- 241 is 432 years; the half life of americium-240 is \(2.12\) days. Why is the 241 isotope a better choice for a smoke detector?

Decay of which nucleus will lead to the following products: (a) bismuth-211 by beta decay; (b) chromium-50 by positron emission; (c) tantalum-179 by electron capture; (d) radium-226 by alpha decay?

A 25.0-mLsample of \(0.050 \mathrm{M}\) barium nitrate solution was mixed with \(25.0 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was \(1.22 \times 10^{6} \mathrm{~Bq} / \mathrm{mL}\). After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of \(250 \mathrm{~Bq} / \mathrm{mL}\). (a) Write a balanced chemical equation for the reaction that occurred. (b) Calculate the \(K_{s p}\) for the precipitate under the conditions of the experiment.

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