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

One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a})_{19}^{39} \mathrm{K}\) and \(_{19}^{40} \mathrm{K},\) \((\mathbf{b})^{209} \mathrm{Bi}\) and \(^{208} \mathrm{Bi}\) \((\mathbf{c})\) nickel-58 and nickel-65.

Short Answer

Expert verified
The radioactive nuclides in each pair are: \(\mathbf{a)}\) \(_{19}^{40} \mathrm{K}\), \(\mathbf{b)}\) \(^{208}\mathrm{Bi}\), and \(\mathbf{c)}\) Nickel-65. This is based on their neutron-proton ratios, which help determine their stability.

Step by step solution

01

Pair A: \(_{19}^{39} \mathrm{K}\) and \(_{19}^{40} \mathrm{K}\)

First, let's find the n/p ratios for both nuclides. For \(_{19}^{39} \mathrm{K}\): - Number of protons (Z) = 19 - Number of neutrons (N) = Mass number (A) - Protons (Z) = 39 - 19 = 20 - n/p ratio = N/Z = 20/19 ≈ 1.05 For \(_{19}^{40} \mathrm{K}\): - Number of protons (Z) = 19 - Number of neutrons (N) = Mass number (A) - Protons (Z) = 40 - 19 = 21 - n/p ratio = N/Z = 21/19 ≈ 1.11 Both nuclides have a similar n/p ratio, but since both are lighter elements (Z ≤ 20), \(_{19}^{39} \mathrm{K}\) with a n/p ratio closer to 1 is more stable. Thus, \(_{19}^{40} \mathrm{K}\) is radioactive.
02

Pair B: \(^{209} \mathrm{Bi}\) and \(^{208}\mathrm{Bi}\)

Let's find the n/p ratios for both nuclides. For \(^{209} \mathrm{Bi}\): - Number of protons (Z) = 83 - Number of neutrons (N) = Mass number (A) - Protons (Z) = 209 - 83 = 126 - n/p ratio = N/Z = 126/83 ≈ 1.52 For \(^{208} \mathrm{Bi}\): - Number of protons (Z) = 83 - Number of neutrons (N) = Mass number (A) - Protons (Z) = 208 - 83 = 125 - n/p ratio = N/Z = 125/83 ≈ 1.51 Both nuclides are heavier elements (Z>20) with a similar n/p ratio. However, \(^{209} \mathrm{Bi}\) is slightly more stable, given its ratio is closer to 1.5; thus, \(^{208}\mathrm{Bi}\) is radioactive.
03

Pair C: Nickel-58 and Nickel-65

Let's find the n/p ratios for both nuclides. For Nickel-58: - Number of protons (Z) = 28 (Nickel atomic number) - Number of neutrons (N) = Mass number (A) - Protons (Z) = 58 - 28 = 30 - n/p ratio = N/Z = 30/28 ≈ 1.07 For Nickel-65: - Number of protons (Z) = 28 (Nickel atomic number) - Number of neutrons (N) = Mass number (A) - Protons (Z) = 65 - 28 = 37 - n/p ratio = N/Z = 37/28 ≈ 1.32 Nickel is a lighter element (Z ≤ 20), so a n/p ratio close to 1 indicates stability. Therefore, Nickel-58 is stable, and Nickel-65 is radioactive.

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.

n/p ratio
The neutron to proton (n/p) ratio is a fundamental aspect in determining the stability of an isotope. It compares the number of neutrons (N) to the number of protons (Z) in an atom's nucleus. The stability of atomic nuclei is closely related to this ratio, with different rules applying based on whether the element is light (with a lower atomic number, Z ≤ 20) or heavy (with a higher atomic number, Z>20).

Stable light elements tend to have n/p ratios close to 1, which means they have nearly equal numbers of neutrons and protons. As elements get heavier, a greater number of neutrons is needed to offset the increasing electrostatic repulsion between the positively charged protons. Therefore, heavier stable isotopes generally have n/p ratios that are increasingly greater than 1, often approaching or exceeding a value around 1.5.

When the n/p ratio deviates significantly from the expected stable range, the nucleide tends to be radioactive, as the forces within the nucleus are not balanced, leading to various modes of radioactive decay to reach a more stable state.
stable isotopes
Stable isotopes are versions of elements that do not undergo radioactive decay over time. They remain unchanged indefinitely. Each element can have one or several stable isotopes, and the key characteristic of these stable isotopes is their n/p ratio, which falls within a certain range that allows the forces within their nuclei to exist in equilibrium.

Stable isotopes are incredibly useful in a variety of scientific applications. For instance, they serve as tracers in biological and geological studies because their non-radioactive nature does not pose a danger to living organisms or the environment. Carbon-12 and Carbon-13 are examples of stable isotopes of carbon, with Carbon-12 being the standard against which atomic weights are measured.
radioactive isotopes
Radioactive isotopes, or radioisotopes, are the unstable forms of elements that emit radiation as they decay into more stable forms. Unlike stable isotopes, radioactive isotopes have n/p ratios that do not align with stability, leading to various modes of decay such as alpha, beta, and gamma decay, each accompanied by the emission of particles or energy.

Radioisotopes have many practical applications, ranging from medical diagnostic and treatment procedures, such as in cancer therapy with isotopes like Technetium-99m, to archaeological dating using radiocarbon dating with Carbon-14. These isotopes are time-sensitive tracers due to their half-lives, which provide scientists with a clock to measure the duration of processes and age of objects. However, the same radioactivity that makes them useful also requires careful handling to prevent unwanted exposure to radiation.

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

Complete and balance the following nuclear equations by supplying the missing particle: \begin{equation}\begin{array}{l}{\text { (a) }_{17}^{14} \mathrm{N}+_{2}^{4} \mathrm{He} \longrightarrow ? +_{1}^{1} \mathrm{H}} \\ {\text { (b) }_{19}^{40} \mathrm{K}+_{1}^{0} \mathrm{e} \ \mathrm{(orbital \ electron) \longrightarrow ?}}\\\ {\text { (c) }_{}{} \mathrm{?}+_{2}^{4} \mathrm{He} \longrightarrow_{14}^{30} \mathrm{Si} +_{1}^{1} \mathrm{H}}\\\ {\text { (d) }_{26}^{58} \mathrm{Fe} +2 _{0}^{1} \mathrm{n} \longrightarrow_{27}^{60} \mathrm{Co}+?}\\\ {\text { (e) }_{92}^{235} \mathrm{U}\longrightarrow+_{0}^{1} n \longrightarrow_{54}^{135} \mathrm{Xe}+2_{0}^{1} \mathrm{n}+?} \end{array}\end{equation}

When two protons fuse in a star, the product is \(^{2} \mathrm{H}\) plus a positron. Write the nuclear equation for this process.

Give the symbol for \((\mathbf{a})\) a proton, \((\mathbf{b})\) a beta particle, \((\mathbf{c})\) a positron.

The naturally occurring radioactive decay series that begins with \(_{92}^{235} \mathrm{U}\) stops with formation of the stable \(_{82}^{207} \mathrm{Pb}\) nucleus. The decays proceed through a series of alpha-particle and beta- particle emissions. How many of each type of emission are involved in this series?

A sample of an alpha emitter having an activity of 0.18 Ci is stored in a 25.0 -mL sealed container at \(22^{\circ} \mathrm{C}\) for 245 days. (a) How many alpha particles are formed during this time? (b) Assuming that each alpha particle is converted to a helium atom, what is the partial pressure of helium gas in the container after this 245 -day period?

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