# 31.6 Binding energy  (Page 4/6)

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## Phet explorations: nuclear fission

Start a chain reaction, or introduce non-radioactive isotopes to prevent one. Control energy production in a nuclear reactor!

## Section summary

• The binding energy (BE) of a nucleus is the energy needed to separate it into individual protons and neutrons. In terms of atomic masses,
$\text{BE}=\left\{\left[\text{Zm}\left({}^{1}\text{H}\right)+{\text{Nm}}_{n}\right]-m\left({}^{A}\text{X}\right)\right\}{c}^{2},$
where $m\left({}^{1}\text{H}\right)$ is the mass of a hydrogen atom, $m\left({}^{A}\text{X}\right)$ is the atomic mass of the nuclide, and ${m}_{n}$ is the mass of a neutron. Patterns in the binding energy per nucleon, $\text{BE}/A$ , reveal details of the nuclear force. The larger the $\text{BE}/A$ , the more stable the nucleus.

## Conceptual questions

Why is the number of neutrons greater than the number of protons in stable nuclei having $A$ greater than about 40, and why is this effect more pronounced for the heaviest nuclei?

## Test prep for ap courses

Binding energy is a measure of how much work must be done against nuclear forces in order to disassemble a nucleus into its constituent parts. For example, the amount of energy in order to disassemble ${}_{2}{}^{4}\text{H}\text{e}$ into 2 protons and 2 neutrons requires 28.3 MeV of work to be done on the nuclear particles. Describe the force that makes it so difficult to pull a nucleus apart. Would it be accurate to say that the electric force plays a role in the forces within a nucleus? Explain why or why not.

## Problems&Exercises

${}^{2}\text{H}$ is a loosely bound isotope of hydrogen. Called deuterium or heavy hydrogen, it is stable but relatively rare—it is 0.015% of natural hydrogen. Note that deuterium has $Z=N$ , which should tend to make it more tightly bound, but both are odd numbers. Calculate $BE/A$ , the binding energy per nucleon, for ${}^{2}\text{H}$ and compare it with the approximate value obtained from the graph in [link] .

1.112 MeV, consistent with graph

${}^{\text{56}}\text{Fe}$ is among the most tightly bound of all nuclides. It is more than 90% of natural iron. Note that ${}^{\text{56}}\text{Fe}$ has even numbers of both protons and neutrons. Calculate $BE/A$ , the binding energy per nucleon, for ${}^{\text{56}}\text{Fe}$ and compare it with the approximate value obtained from the graph in [link] .

${}^{\text{209}}\text{Bi}$ is the heaviest stable nuclide, and its $\text{BE}/A$ is low compared with medium-mass nuclides. Calculate $BE/A$ , the binding energy per nucleon, for ${}^{\text{209}}\text{Bi}$ and compare it with the approximate value obtained from the graph in [link] .

7.848 MeV, consistent with graph

(a) Calculate $\text{BE}/A$ for ${}^{\text{235}}\text{U}$ , the rarer of the two most common uranium isotopes. (b) Calculate $\text{BE}/A$ for ${}^{\text{238}}\text{U}$ . (Most of uranium is ${}^{\text{238}}\text{U}$ .) Note that ${}^{\text{238}}\text{U}$ has even numbers of both protons and neutrons. Is the $\text{BE}/A$ of ${}^{\text{238}}\text{U}$ significantly different from that of ${}^{\text{235}}\text{U}$ ?

(a) Calculate $\text{BE}/A$ for ${}^{\text{12}}\text{C}$ . Stable and relatively tightly bound, this nuclide is most of natural carbon. (b) Calculate $\text{BE}/A$ for ${}^{\text{14}}\text{C}$ . Is the difference in $\text{BE}/A$ between ${}^{\text{12}}\text{C}$ and ${}^{\text{14}}\text{C}$ significant? One is stable and common, and the other is unstable and rare.

(a) 7.680 MeV, consistent with graph

(b) 7.520 MeV, consistent with graph. Not significantly different from value for ${}^{\text{12}}\text{C}$ , but sufficiently lower to allow decay into another nuclide that is more tightly bound.

The fact that $\text{BE}/A$ is greatest for $A$ near 60 implies that the range of the nuclear force is about the diameter of such nuclides. (a) Calculate the diameter of an $A=\text{60}$ nucleus. (b) Compare $\text{BE}/A$ for ${}^{\text{58}}\text{Ni}$ and ${}^{\text{90}}\text{Sr}$ . The first is one of the most tightly bound nuclides, while the second is larger and less tightly bound.

The purpose of this problem is to show in three ways that the binding energy of the electron in a hydrogen atom is negligible compared with the masses of the proton and electron. (a) Calculate the mass equivalent in u of the 13.6-eV binding energy of an electron in a hydrogen atom, and compare this with the mass of the hydrogen atom obtained from [link] . (b) Subtract the mass of the proton given in [link] from the mass of the hydrogen atom given in [link] . You will find the difference is equal to the electron’s mass to three digits, implying the binding energy is small in comparison. (c) Take the ratio of the binding energy of the electron (13.6 eV) to the energy equivalent of the electron’s mass (0.511 MeV). (d) Discuss how your answers confirm the stated purpose of this problem.

(a) $1\text{.}\text{46}×{\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}u$ vs. 1.007825 u for ${}^{1}\text{H}$

(b) 0.000549 u

(c) $2\text{.}\text{66}×{\text{10}}^{-5}$

Unreasonable Results

A particle physicist discovers a neutral particle with a mass of 2.02733 u that he assumes is two neutrons bound together. (a) Find the binding energy. (b) What is unreasonable about this result? (c) What assumptions are unreasonable or inconsistent?

(a) $–9.315 MeV$

(b) The negative binding energy implies an unbound system.

(c) This assumption that it is two bound neutrons is incorrect.

#### Questions & Answers

Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have?
how would I work this problem
Alexia
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
Igor
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