/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 The atomic masses of hydrogen-2 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The atomic masses of hydrogen-2 (deuterium), helium-4, and lithium-6 are 2.014102 amu, 4.002602 amu, and 6.0151228 amu, respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon. (d) Which of these three isotopes has the largest nuclear binding energy per nucleon? Does this agree with the trends plotted in Figure 21.12\(?\)

Short Answer

Expert verified
The nuclear masses of hydrogen-2 (deuterium), helium-4, and lithium-6 are 3.34399 x 10^-27 kg, 6.64412 x 10^-27 kg, and 9.98864 x 10^-27 kg, respectively. The nuclear binding energies are 2.224 x 10^-13 J, 4.0323 x 10^-12 J, and 3.2994 x 10^-12 J, respectively. The nuclear binding energy per nucleon for each isotope is 1.112 x 10^-13 J/nucleon, 1.008075 x 10^-12 J/nucleon, and 5.499 x 10^-13 J/nucleon, respectively. Helium-4 has the largest nuclear binding energy per nucleon, with a value of 1.008075 x 10^-12 J/nucleon. This result should agree with the trends observed in Figure 21.12.

Step by step solution

01

Calculate the Nuclear Mass for Each Isotope

First, let's find the nuclear masses for each isotope. We can do this by multiplying the atomic mass units (amu) by the conversion factor to get the mass in kg. Conversion factor: 1 amu = 1.66054 x 10^-27 kg For hydrogen-2 (deuterium): Atomic mass = 2.014102 amu Nuclear mass = 2.014102 amu * (1.66054 x 10^-27 kg/amu) = 3.34399 x 10^-27 kg For helium-4: Atomic mass = 4.002602 amu Nuclear mass = 4.002602 amu * (1.66054 x 10^-27 kg/amu) = 6.64412 x 10^-27 kg For lithium-6: Atomic mass = 6.0151228 amu Nuclear mass = 6.0151228 amu * (1.66054 x 10^-27 kg/amu) = 9.98864 x 10^-27 kg
02

Calculate the Nuclear Binding Energy for Each Isotope

Next, we'll calculate the nuclear binding energy for each isotope using the formula: Nuclear binding energy = (Z * mass_H + N * mass_n - nuclear_mass) * c^2 where Z: number of protons, mass_H: mass of a proton (1.007825 amu), N: number of neutrons, mass_n: mass of a neutron (1.008665 amu), c: speed of light (3 x 10^8 m/s) For hydrogen-2: Z = 1, N = 1 Binding energy = [(1 * 1.007825) + (1 * 1.008665) - 2.014102] * (1.66054 x 10^-27 kg/amu) * (3 x 10^8 m/s)^2 Binding energy = 2.224 x 10^-13 J For helium-4: Z = 2, N = 2 Binding energy = [(2 * 1.007825) + (2 * 1.008665) - 4.002602] * (1.66054 x 10^-27 kg/amu) * (3 x 10^8 m/s)^2 Binding energy = 4.0323 x 10^-12 J For lithium-6: Z = 3, N = 3 Binding energy = [(3 * 1.007825) + (3 * 1.008665) - 6.0151228] * (1.66054 x 10^-27 kg/amu) * (3 x 10^8 m/s)^2 Binding energy = 3.2994 x 10^-12 J
03

Calculate the Nuclear Binding Energy per Nucleon for Each Isotope

Now, we'll find the nuclear binding energy per nucleon for each isotope by dividing the nuclear binding energy by the number of nucleons (protons + neutrons). For hydrogen-2: Binding energy per nucleon = (2.224 x 10^-13 J) / 2 = 1.112 x 10^-13 J/nucleon For helium-4: Binding energy per nucleon = (4.0323 x 10^-12 J) / 4 = 1.008075 x 10^-12 J/nucleon For lithium-6: Binding energy per nucleon = (3.2994 x 10^-12 J) / 6 = 5.499 x 10^-13 J/nucleon
04

Determine the Isotope with the Largest Nuclear Binding Energy per Nucleon

Comparing the nuclear binding energy per nucleon for each isotope: Hydrogen-2: 1.112 x 10^-13 J/nucleon Helium-4: 1.008075 x 10^-12 J/nucleon Lithium-6: 5.499 x 10^-13 J/nucleon Helium-4 has the largest nuclear binding energy per nucleon, with a value of 1.008075 x 10^-12 J/nucleon.
05

Check against Figure 21.12

We're unable to view Figure 21.12, but the general trend typically shows an increase in nuclear binding energy per nucleon as the atomic number increases, reaching a peak around the atomic number of iron. Our result of helium-4 having the largest binding energy per nucleon among the three given isotopes is expected based on its higher atomic number compared to hydrogen-2. For the lithium-6 isotope, it doesn't follow the general trend for lighter elements because its atomic mass is relatively higher than helium-4. This may cause some deviations, which might be evident in Figure 21.12.

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 91Ó°ÊÓ!

Key Concepts

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

Atomic Mass
The atomic mass of an isotope is its total mass, measured in atomic mass units (amu). This mass includes the combined mass of protons, neutrons, and electrons in an atom. However, since electrons have a comparatively tiny mass, the atomic mass is largely determined by the nucleus, composed of protons and neutrons.
Atomic mass gives us a sense of how much the isotope weighs compared to carbon-12, the standard reference, which is assigned exactly 12 amu. This measurement aids in calculating nuclear properties like binding energy.
To convert atomic mass from amu to kilograms, use the conversion factor:
  • 1 amu = 1.66054 x 10^-27 kg
Calculating the nuclear mass in these units is crucial, as it serves as a basis for understanding the energy dynamics within the nucleus.
Isotopes
Isotopes are different forms of the same chemical element. They have the same number of protons but differ in the number of neutrons within the nucleus. This difference results in distinct atomic masses for each isotope.
Despite having different masses, isotopes of an element exhibit similar chemical behavior since they have the same electronic structure. For example, hydrogen-2 (deuterium) and hydrogen-1 (protium) both react similarly chemically but have different physical properties.
Exploring nuclear properties such as binding energy helps understand why some isotopes are more stable than others. Isotopes are named based on their mass number, calculated as the sum of protons and neutrons. This understanding is crucial for fields like nuclear physics and chemistry.
Knowledge of isotopes is essential when evaluating nuclear reactions or decay processes, where different isotopes can lead to vastly different results.
Mass-Energy Equivalence
The principle of mass-energy equivalence is summed up by Einstein’s famous equation, \( E = mc^2 \). This concept implies that mass can be converted into energy, and vice versa. It plays a pivotal role in calculating nuclear binding energy.
In the context of nuclear physics, when a nucleus forms, a small fraction of its mass is converted into energy, which holds the nucleus together. This energy is the nuclear binding energy — the energy required to disassemble the nucleus into individual protons and neutrons.
Mass-energy equivalence also helps to calculate the nuclear binding energy per nucleon, which is an indicator of the stability of an isotope. Higher binding energy per nucleon generally means a more stable nucleus.
  • Formula for nuclear binding energy: \( \text{Binding Energy} = (\text{mass difference}) \times c^2 \)
  • Speed of light \( c = 3 \times 10^8 \) m/s
By understanding mass-energy equivalence, students can grasp how energy conservation occurs at a subatomic level, emphasizing the interconnectivity of mass and energy in the universe.

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

Based on the following atomic mass values \(-^{1} \mathrm{H}, 1.00782\) \(\mathrm{amu} ;^{2} \mathrm{H}, 2.01410 \mathrm{amu}\); \(^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;^{3} \mathrm{He}, 3.01603\) \(\mathrm{amu} ;^{4} \mathrm{He}, 4.00260 \mathrm{amu}-\) amu—and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: \begin{equation}(\mathbf{a})\quad_{1}^{2} \mathrm{H}+_{1}^{3} \mathrm{H} \longrightarrow _{4}^{2} \mathrm{He}+_{1}^{0} \mathrm{n}\end{equation} \begin{equation}(\mathbf{b})\quad_{1}^{2} \mathrm{H}+_{1}^{2} \mathrm{H} \longrightarrow_{2}^{3} \mathrm{He}+_{0}^{1} \mathrm{n}\end{equation} \begin{equation}(\mathbf{c})\quad_{1}^{2} \mathrm{H}+_{2}^{3} \mathrm{He} \longrightarrow_{2}^{4} \mathrm{He}+_{1}^{1} \mathrm{H}\end{equation}

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a})_{20}^{40} \mathrm{Ca}\) and \(_{20}^{45} \mathrm{Ca}\) , \((\mathbf{b})^{12} \mathrm{C}\) and \(^{14} \mathrm{C},\) \((\mathbf{c})\) lead-206 and thorium-230. Explain your choice in each case.

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?

Which of the following statements about the uranium used in nuclear reactors is or are true? (i) Natural uranium has too little \(^{235} U\) to be used as a fuel. (ii) \(^{238} U\) cannot be used as a fuel because it forms a supercritical mass too easily. (iii) To be used as fuel, uranium must be enriched so that it is more than 50\(\%^{235} \mathrm{U}\) in composition. (iv) The neutron-induced fission of \(^{235} \mathrm{U}\) releases more neutrons per nucleus than fission of \(^{238} \mathrm{U}\).

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.