Chapter 16: Problem 56

The tantalum isotope \({ }_{73}^{186} \mathrm{Ta}\) is radioactive and decays by converting a neutron to a proton. (a) Where is this atom likely to lie in the band of stability? (b) Write a nuclear reaction for this decay process. (c) Which type of decay is this?

Short Answer

Expert verified

The tantalum isotope \({ }_{73}^{186}\mathrm{Ta}\) lies on the neutron-rich side of the band of stability, with a neutron-to-proton ratio of approximately 1.55. The decay process can be represented by the nuclear reaction: \[^{186}_{73}\mathrm{Ta} \rightarrow ^{186}_{74}\mathrm{W} + e^{-} + \bar{\nu}_e\], where a neutron converts into a proton, releasing an electron and an electron antineutrino. This process is known as beta-minus (\(\beta^{-}\)) decay.

Step by step solution

Determine the position of the isotope in the band of stability

In order to determine where the tantalum isotope \({ }_{73}^{186}\mathrm{Ta}\) is likely to lie in the band of stability, we need to consider its proton-to-neutron ratio. The band of stability is the region in which stable isotopes have the right proton-to-neutron ratio, which is approximately 1:1 for light nuclei (Z<20) and shifts to a higher neutron-to-proton ratio for heavier nuclei. Since the given tantalum isotope has 73 protons and 186 - 73 = 113 neutrons, its neutron-to-proton ratio is approximately 113/73 ≈ 1.55. This is higher than the 1:1 ratio for light elements, so it is likely to lie on the neutron-rich side of the band of stability, as expected for heavier nuclei.

Write a nuclear reaction for the decay process

We are given that the tantalum isotope \({ }_{73}^{186}\mathrm{Ta}\) decays by converting a neutron to a proton. This decay process can be represented by the following nuclear reaction: \[^{186}_{73}\mathrm{Ta} \rightarrow ^{186}_{74}\mathrm{W} + e^{-} + \bar{\nu}_e\] A neutron in the nucleus of the tantalum atom converts into a proton, releasing an electron \(e^{-}\) (also known as a beta particle) and an electron antineutrino \(\bar{\nu}_e\). This process results in a new isotope, \({ }_{74}^{186}\mathrm{W}\), also known as tungsten-186, which has one more proton and one less neutron than the original tantalum isotope.

Identify the type of decay

The decay process in which a neutron is converted into a proton, releasing an electron and an electron antineutrino, is known as beta-minus (\(\beta^{-}\)) decay. Therefore, the tantalum isotope \({ }_{73}^{186}\mathrm{Ta}\) decays by \(\beta^{-}\) decay.

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!

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

The tantalum isotope \({ }_{73}^{186} \mathrm{Ta}\) is radioactive and decays by converting a neutron to a proton. (a) Where is this atom likely to lie in the band of stability? (b) Write a nuclear reaction for this decay process. (c) Which type of decay is this?

Short Answer

Step by step solution

Determine the position of the isotope in the band of stability

Write a nuclear reaction for the decay process

Identify the type of decay

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Chemical Reactions

Organic Chemistry

Ionic and Molecular Compounds

Nuclear Chemistry

Chemistry Branches

Study anywhere. Anytime. Across all devices.