Problem 48 - Chemistry The Central Science

Question

The atomic masses of nitrogen- 14 , titanium- 48 , and xenon129 are 13.999234 amu, 47.935878 amu, and 128.904779 amu, respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon.

Step-by-Step Solution

Answer

For the given isotopes, we have calculated (a) the Nuclear Masses: Nitrogen-14: 13.999234 amu, Titanium-48: 47.935878 amu, Xenon-129: 128.904779 amu; (b) Nuclear Binding Energies: Nitrogen-14: 127.0 MeV, Titanium-48: 324.3 MeV, Xenon-129: 754.3 MeV; and (c) Nuclear Binding Energy per Nucleon: Nitrogen-14: 9.07 MeV/Nucleon, Titanium-48: 6.76 MeV/Nucleon, Xenon-129: 5.85 MeV/Nucleon.

Step 1: Calculate Nuclear Mass

For each isotope, the nuclear mass (M) can be calculated using the atomic mass number (A) and the atomic mass units (amu) provided. For Nitrogen-14: M = 13.999234 amu For Titanium-48: M = 47.935878 amu For Xenon-129: M = 128.904779 amu

Step 2: Calculate Nuclear Binding Energy

First, let's find the mass defect (∆m) for each isotope, using the atomic mass number (A) and the mass of protons and neutrons: Nitrogen-14: Finding mass defect (∆m): Number of protons = 7 Number of neutrons = 7 mass of proton = 1.007276 amu mass of neutron = 1.008665 amu ∆m = 7*(1.007276 amu) + 7*(1.008665 amu) - 13.999234 amu ∆m = 0.136332 amu Binding energy (BE) = ∆m * c² Using \(c = 931.5\frac{MeV}{c². amu}\) BE = 0.136332 amu * 931.5 MeV/c².amu = 127.0 MeV Titanium-48: Finding mass defect (∆m): Number of protons = 22 Number of neutrons = 26 ∆m = 22*(1.007276 amu) + 26*(1.008665 amu) - 47.935878 amu ∆m = 0.348219 amu Binding energy (BE) = ∆m * c² BE = 0.348219 amu * 931.5 MeV/c².amu = 324.3 MeV Xenon-129: Finding mass defect (∆m): Number of protons = 54 Number of neutrons = 75 ∆m = 54*(1.007276 amu) + 75*(1.008665 amu) - 128.904779 amu ∆m = 0.809360 amu Binding energy (BE) = ∆m * c² BE = 0.809360 amu * 931.5 MeV/c².amu = 754.3 MeV

Step 3: Calculate Nuclear Binding Energy per Nucleon

Now, we can calculate the nuclear binding energy per nucleon by dividing the total nuclear binding energy by the atomic mass number (A). For Nitrogen-14: Nuclear binding energy per nucleon = \(\frac{127.0\, MeV}{14}\) = 9.07 MeV/Nucleon For Titanium-48: Nuclear binding energy per nucleon = \(\frac{324.3\, MeV}{48}\) = 6.76 MeV/Nucleon For Xenon-129: Nuclear binding energy per nucleon = \(\frac{754.3\, MeV}{129}\) = 5.85 MeV/Nucleon

Key Concepts

Nuclear Mass

Understanding nuclear mass is crucial to grasp many aspects of nuclear physics. The nuclear mass refers to the total mass of the protons and neutrons (nucleons) within an atom's nucleus. Normally, we approximate the mass of protons and neutrons as 1 atomic mass unit (amu) each. However, this is not entirely accurate because protons and neutrons themselves are made up of quarks and gluons, whose interactions also contribute to the total nuclear mass.

To calculate nuclear mass, as in our exercise, we simply take the sum of the protons' mass and neutrons' mass and compare it with the mass of the atom as given on the periodic table (or here as part of the problem statement). This concept is essential not only in pure physics but also has implications in other areas like nuclear medicine and energy generation.

Mass Defect

Now, let's delve into the concept of mass defect. Mass defect is the difference between the expected mass of the nucleus (if protons and neutrons were added together without any binding energy) and its actual mass. If we were to sum the individual masses of protons and neutrons in a nucleus and then subtract the actual nuclear mass, we would detect a small amount of mass missing. This missing mass is what we call the mass defect.

It's intriguing to note that this 'defect' doesn’t mean something is wrong; rather, it is a consequence of energy release when nucleons bind together. This is illustrated by Albert Einstein’s famous equation, E=mc², which tells us that this 'missing' mass has been converted into binding energy that holds the nucleus together. Instantly solving this part of the puzzle helps us understand why nuclei are stable and what makes them tick.

Binding Energy per Nucleon

Lastly, to complete our nuclear trilogy of concepts, there's the binding energy per nucleon. This is an average energy value typically used to represent the stability of a nucleus. To calculate this, take the nuclear binding energy (which is the energy derived from the mass defect) and divide it by the number of nucleons in the nucleus.

The resulting number tells us how strongly each nucleon is held in the nucleus. Generally, the higher the binding energy per nucleon, the more stable the nucleus. In our textbook example, we calculate this for various elements, noticing that the value differs from one element to another. This helps explain why some elements are more prone to undergo nuclear reactions than others, which is crucial for topics like nuclear fission and fusion, both of which play a role in energy production and cosmic events.