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Chapter 21: Problem 54

The isotope \({ }_{28}^{62} \mathrm{Ni}\) has the largest binding energy per nucleon of any isotope. Calculate this value from the atomic mass of nickel-62 \((61.928345 \mathrm{u})\) and compare it with the value given for iron- 56 in Table 21.7 .

Short Answer

Expert verified
The binding energy per nucleon for nickel-62 can be calculated using the formula: \(Binding \ Energy = (Z * m_p + (A - Z) * m_n - M) * c^2\), where Z = 28, A = 62, m_p = 1.007276 u, m_n = 1.008664 u, and M = 61.928345 u. After calculating the binding energy, we can find the binding energy per nucleon by dividing it by the total number of nucleons (A). Finally, compare this value to the binding energy per nucleon for iron-56 given in Table 21.7.

Step by step solution

01

Recall the binding energy formula and the relevant constants

To calculate the binding energy, we need to use the formula: Binding Energy = (Z * m_p + (A - Z) * m_n - M) * c^2 where Z = number of protons, A = number of nucleons, m_p = mass of a proton (1.007276 u), m_n = mass of a neutron (1.008664 u), M = atomic mass of the isotope (given as 61.928345 u for nickel-62), c = speed of light (approximately 2.99792458 * 10^8 m/s, but we will only require the square of c which is c^2 = 8.98755179 * 10^16 m^2/s^2), and 1 u = 931.494 MeV/c^2.
02

Calculate the number of protons and neutrons for nickel-62

The isotope notation indicates that nickel-62 has an atomic number (Z) of 28, which means it has 28 protons. To find the number of neutrons (A - Z), we subtract the number of protons from the total number of nucleons, which is also given in the isotope notation as 62. Number of neutrons = A - Z = 62 - 28 = 34
03

Calculate the binding energy of nickel-62

Now that we have all the relevant values, we can plug them into the binding energy formula: Binding Energy = (Z * m_p + (A - Z) * m_n - M) * c^2 = (28 * 1.007276 + 34 * 1.008664 - 61.928345) * 8.98755179 * 10^16 * (931.494)
04

Calculate the binding energy per nucleon for nickel-62

To find the binding energy per nucleon, divide the binding energy by the total number of nucleons (A): Binding Energy per nucleon = Binding Energy / A
05

Compare the binding energy per nucleon of nickel-62 with that of iron-56

Refer to Table 21.7 provided for the binding energy per nucleon of iron-56. Compare the calculated binding energy per nucleon of nickel-62 with this value.

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Key Concepts

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

Nucleons
In the nucleus of an atom, nucleons are the particles that bind together thanks to the strong force. They consist of protons, which are positively charged, and neutrons, which have no charge. The number of nucleons in an atom is crucial because it defines the mass number, represented as \(A\).
Nickel-62, for example, has 62 nucleons. This means a combination of protons and neutrons that add up to this number. Knowing the number of protons (atomic number, \(Z\)) allows you to find the number of neutrons by subtracting \(Z\) from \(A\).
  • Protons: Determine the element's chemical properties.
  • Neutrons: Affect the mass and stability of an atom.
Understanding nucleons helps to calculate the binding energy, which measures how strongly these particles are held together. A greater binding energy per nucleon often indicates more nuclear stability.
Isotopes
Isotopes are variations of the same element, having the same number of protons but different numbers of neutrons. This means they have different mass numbers. For instance, nickel-62 is one isotope among other isotopes of nickel. All isotopes of nickel have 28 protons, but nickel-62 differs due to its 34 neutrons.
Isotopes are fascinating because they often have similar chemical behavior but distinct nuclear properties. The stability of an isotope can vary significantly, affecting its abundance and use in various applications, such as dating geological samples or medical imaging.
  • Stable isotopes: Do not undergo radioactive decay.
  • Radioactive isotopes: Decay over time, releasing particles and energy.
The concept of isotopes is pivotal in nuclear chemistry and physics, influencing our understanding of elements and allowing us to harness nuclear energy.
Atomic Mass
Atomic mass, often denoted as \(M\) in formulas, is the mass of an atom expressed in atomic mass units (u). It's crucial for understanding atomic and nuclear properties. Atomic mass includes both protons and neutrons within the nucleus, while electrons have negligible mass in this context.
Nickel-62 has an atomic mass of 61.928345 u, indicating the total mass of its protons and neutrons. Atomic mass is essential for calculating binding energy, which requires precise values.
Here's why atomic mass is important:
  • Determines the element's characteristics and stability.
  • Vital for calculations involving nuclear reactions and processes.
Atomic mass varies slightly between isotopes, contributing to their unique identities and uses in science and technology.

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Most popular questions from this chapter

Which of the following statements best explains why alpha emission is relatively common, but proton emission is extremely rare? (a) Alpha particles are very stable because of magic numbers of protons and neutrons. (b) Alpha particles occur in the nucleus. (c) Alpha particles are the nuclei of an inert gas. (d) An alpha particle has a higher charge than a proton.

(a) Which of the following are required characteristics of an isotope to be used as a fuel in a nuclear power reactor? (i) It must emit gamma radiation. (ii) On decay, it must release two or more neutrons. (iii) It must have a half-life less than one hour. (iv) It must undergo fission upon the absorption of a neutron. (b) What is the most common fissionable isotope in a commercial nuclear power reactor?

Write balanced nuclear equations for the following processes: (a) radon-198 undergoes alpha emission; (b) thorium-234 undergoes beta emission; (c) copper-61 undergoes positron emission; (d) silver-106 undergoes electron capture.

Despite the similarities in the chemical reactivity of elements in the lanthanide series, their abundances in Earth's crust vary by two orders of magnitude. This graph shows the relative abundance as a function of atomic number. Which of the following statements best explains the sawtooth variation across the series? (a) The elements with an odd atomic number lie above the belt of stability. (b) The elements with an odd atomic number lie below the belt of stability. (c) The elements with an even atomic number have a magic number of protons. (d) Pairs of protons have a special stability.

The Sun radiates energy into space at the rate of \(3.9 \times 10^{26} \mathrm{~J} / \mathrm{s} .\) (a) Calculate the rate of mass loss from the Sun in kg/s. (b) How does this mass loss arise? (c) It is estimated that the Sun contains \(9 \times 10^{56}\) free protons. How many protons per second are consumed in nuclear reactions in the Sun?
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