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Calculate the mass defect, the binding energy (in \(\mathrm{MeV}\) ), and the binding energy per nucleon of (a) the nitrogen nucleus, \({ }_{7}^{14} \mathrm{~N},\) and (b) the helium nucleus, \({ }_{2}^{4} \mathrm{He}\). (c) How does the binding energy per nucleon compare for these two nuclei?

Short Answer

Expert verified
To calculate the mass defect, subtract the total mass of the nucleons in the nucleus from the actual mass of the nucleus. Then, multiply this defect with \(931.5\, MeV\) to obtain the total binding energy. Dividing it by the number of nucleons gives the binding energy per nucleon. By comparing these values for \(_{7}^{14}\mathrm{N}\) and \(_{2}^{4}\mathrm{He}\), it can be concluded which of them are more tightly bound. For the exact values, calculations should be executed with accurate mass values available from the lookup table of isotopes.

Step by step solution

01

Calculate the Mass Defect

To calculate the mass defect, subtract the total mass of the nucleons in the nucleus from the actual mass of the nucleus. The actual mass of a nucleus can be found in a table of isotopes. The masses to consider are: 1 neutron = 1.008665 u and 1 proton = 1.007825 u. For \(_{7}^{14}\mathrm{N}\), the mass defect = \(14 \, u - 7 \times 1.007825\, u - 7 \times 1.008665\, u\). For \(_{2}^{4}\mathrm{He}\), the mass defect = \(4 \, u - 2 \times 1.007825\, u - 2 \times 1.008665\, u\).
02

Calculate the Binding Energy

The binding energy is the mass defect converted into energy via Einstein's equation, \(E = mc^2\). Consider that 1 u of mass defect equals \(931.5\, MeV\). So, the binding energy for \(_{7}^{14}\mathrm{N}\) and \(_{2}^{4}\mathrm{He}\) can be calculated by multiplying their respective mass defects by \(931.5\, MeV\).
03

Calculate the Binding Energy per Nucleon

To calculate the binding energy per nucleon, divide the total binding energy by the number of nucleons (protons and neutrons). For \(_{7}^{14}\mathrm{N}\), the number of nucleons is \(14\), and for \(_{2}^{4}\mathrm{He}\), the number of nucleons is \(4\).
04

Compare the Binding Energy per Nucleon

With the binding energy per nucleon calculated for each of \(_{7}^{14}\mathrm{N}\) and \(_{2}^{4}\mathrm{He}\), one can compare by reviewing the absolute values.

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Key Concepts

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

Mass Defect
Mass defect is a critical notion in the realm of nuclear physics, playing a pivotal role in understanding the binding energy of a nucleus. It can be somewhat perplexing, but let's simplify it. Imagine assembling a puzzle; the individual pieces symbolize protons and neutrons. The completed puzzle signifies the nucleus. Now, curiously, when you weigh the completed puzzle, it's lighter than the total weight of the individual pieces. This weight difference is what we call the 'mass defect'.

Mass defect is pivotal because it exhibits the mass converted into energy to bind the nucleons together in the nucleus. It's not lost but rather transformed in line with the concept of mass-energy equivalence. The calculation begins by noting the individual masses of protons and neutrons, then measuring the mass of the nucleus itself (which can often be found in tables of isotopic masses). The mass defect is the disparity between the total mass of protons and neutrons when separate and when they are part of the nucleus.
Nucleus Mass
When discussing nuclear physics, it's important to understand that the 'nucleus mass' refers to the actual, observable mass of a nucleus of an atom. Contrary to the summed mass of its individual protons and neutrons (also termed nucleons), the nucleus mass, as previously noted with mass defect, is invariably less due to the binding energy involved.

The nucleus mass can be measured precisely using instruments like mass spectrometers. For the calculation of binding energy, accurate figures of the nucleus mass are paramount. In the case of textbook problems, these masses are typically provided. This differentiated mass from the sum of its constituent nucleons is proof of the binding energy's manifestation and is foundational to the stability of every atomic nucleus you encounter in nature.
Einstein's Equation E=mc^2
Einstein's revolutionary equation, E=mc^2, is a cornerstone of modern physics, illustrating the interchangeability of mass and energy. This equation reveals that a tiny amount of mass can be converted into a significant amount of energy and vice versa. The 'E' stands for energy, 'm' for mass, and 'c' is the speed of light in a vacuum (about 299,792,458 meters per second).

For binding energy calculations, this equation is the magic formula that allows us to convert the mass defect into energy units, typically in megaelectronvolts (MeV). It explains why the nucleus mass being less than the combined mass of protons and neutrons results in a release of energy, which is the binding energy stabilizing the nucleus. Clearly understanding this equation is vital for students, as it is the turning point of many critical calculations in nuclear physics.
Binding Energy per Nucleon
Binding energy per nucleon reveals a lot about the nucleus' stability. It is the quotient of the total binding energy of a nucleus and the total number of nucleons it contains. To digest this, consider it as a benchmark for stability—the higher the binding energy per nucleon, the more stable the nucleus. This concept is meaningful across various elements and isotopes, allowing scientists to compare which nuclei are more tightly bound.

When it's time to crunch the numbers, this ratio helps determine whether a nucleus will tend to bind together or if it might split apart or fuse with another nucleus - key to understanding nuclear reactions like fusion and fission. By diving into this measure, students gain insights into the relative strength of nuclear bonds across different elements. This is where the energy that powers stars, nuclear reactors, and even atomic bombs comes from. It's worth noting that the energy trends across the periodic table generally show an increase in binding energy per nucleon up to iron (Fe), beyond which it gradually decreases.

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Most popular questions from this chapter

An alpha particle is strongly bound. The \({ }_{6}^{12} \mathrm{C}\) nucleus might be modeled as a composite of three alpha particles. Compare the binding energy of \({ }_{6}^{12} \mathrm{C}\) with three times the binding energy of an alpha particle. Which of these quantities is larger, and why might this be so?

The common isotope of uranium, \({ }^{238} \mathrm{U},\) has a half-life of \(4.47 \times 10^{9}\) years, decaying to \({ }^{234} \mathrm{Th}\) by alpha emission. (a) What is the decay constant? (b) What mass of uranium is required for an activity of 1.00 curie? (c) How many alpha particles are emitted per second by \(10.0 \mathrm{~g}\) of uranium?

A sample of radioactive nuclei has \(N_{0}\) nuclei at time \(t=0 .\) The half- life of the decay is \(T_{1 / 2}\). In terms of \(N_{0}\), how many decays occur in the time period between \(t=0\) and \(t=0.500 T_{1 / 2} ?\)

Radioactive Tracers. Radioactive isotopes are often introduced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. One such tracer is \({ }^{131} \mathrm{I}\), a \(\beta^{-}\) emitter with a half-life of \(8.0 \mathrm{~d}\). Suppose a scientist introduces a sample with an activity of \(325 \mathrm{~Bq}\) and watches it spread to the organs. (a) Assuming that all of the sample went to the thyroid gland, what will be the decay rate in that gland 24 d (about \(3 \frac{1}{2}\) weeks) later? (b) If the decay rate in the thyroid 24 d later is measured to be 17.0 Bq, what percentage of the tracer went to that gland? (c) What isotope remains after the I-131 decays?

Tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) is an unstable isotope of hydrogen; its mass, including one electron, is 3.016049 u. (a) Show that tritium must be unstable with respect to beta decay because the decay products ( \({ }_{2}^{3}\) He plus an emitted electron) have less total mass than the tritium. (b) Determine the total kinetic energy (in MeV) of the decay products, taking care to account for the electron masses correctly.

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