Question Video: Determining the Binding Energy per Nucleon for an Atom of Lithium-7 | Nagwa Question Video: Determining the Binding Energy per Nucleon for an Atom of Lithium-7 | Nagwa

# Question Video: Determining the Binding Energy per Nucleon for an Atom of Lithium-7 Chemistry • First Year of Secondary School

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What is the average binding energy per nucleon in units of mega-electron volts for an atom of lithium-7 with an observed mass of 7.01435 u? Give your answer to 2 decimal places. Take the masses of a proton and a neutron to be 1.00728 u and 1.00866 u respectively.

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### Video Transcript

What is the average binding energy per nucleon in units of megaelectron volts for an atom of lithium-7 with an observed mass of 7.01435 unified atomic mass units? Give your answer to two decimal places. Take the masses of a proton and a neutron to be 1.00728 unified atomic mass units and 1.00866 unified atomic mass units, respectively.

Binding energy is the energy required to disassemble the nucleus of an atom into its constituent nucleons, where a nucleon is a particle in the nucleus. It can be a proton or a neutron. A significant amount of energy is required to break a nucleus into its constituent nucleons. Or in the case of the reverse process, a significant amount of energy is released upon the formation of a nucleus. The relationship between energy and mass is described by Einsteinโs equation ๐ธ equals ๐๐ squared, where ๐ธ represents energy, ๐ is mass, and ๐ is the constant, the speed of light.

As the amount of energy released is large, it has a significant mass. This means that the mass of a nucleus is always less than the sum of the individual masses of its constituent nucleons. The difference between these masses is called the mass defect. And it is equivalent to the binding energy of the nucleus. The question asks us to find the binding energy. The binding energy is equivalent to the mass defect. So our first step is to calculate the mass defect. The mass defect is the combined mass of nucleons minus the mass of the nucleus, which we call the observed mass. Weโve been given the masses of a proton and a neutron in the question, but we first need to calculate how many of each we have.

The question tells us we have lithium-7. If we find lithium in the periodic table, we see that the atomic number or proton number is three. So we know that lithium has three protons. The isotope we have is lithium-7, where seven represents the mass number of the isotope and is the total number of protons and neutrons. So to find the number of neutrons, we subtract the number of protons, which is three, from the total number of protons and neutrons, which is seven. This gives us a value of four. So in lithium-7, there are three protons and four neutrons.

To find the combined mass of nucleons, we need to multiply the number of protons, which is three, by the mass of a proton, which we are told in the question is 1.00728 unified atomic mass units. We then need to add the number of neutrons, which is four, multiplied by the mass of a neutron, which we are told in the question is 1.00866 unified atomic mass units. We then need to subtract the observed mass, which we are told in the question is 7.01435 unified atomic mass units. If we perform this calculation, we get a value of 0.04213 unified atomic mass units.

Our next step is to convert from unified atomic mass units to kilograms. We can use the conversion that one unified atomic mass unit is equivalent to 1.66 times 10 to the power of minus 27 kilograms. As we want to end up with the units in kilograms, we need to multiply the mass defect in unified atomic mass units by 1.66 times 10 to the power of minus 27 kilograms and then divide this value by one unified atomic mass unit as that is what it is equivalent to. The unified atomic mass units cancel. And we get a value of 6.99358 times 10 to the power of minus 29 kilograms.

Now that we have the mass defect in kilograms, we can use ๐ธ equals ๐๐ squared to calculate the binding energy. In this equation, all units are SI units. The energy is in joules, the mass is in kilograms, and the speed of light is in meters per second. We are trying to find ๐ธ, the energy, in joules, and we know the mass to be 6.99 times 10 to the power of minus 29 kilograms. We need to multiply the mass by the speed of light squared, where the speed of light is three times 10 to the power of eight meters per second. If we perform the calculation, we get a value of 6.294222 times 10 to the power of minus 12 joules.

So we have now calculated the binding energy. But the question asks for it in units of megaelectron volts. So our next step is to convert the units from joules to megaelectron volts. We can use the conversion that one electron volt is equivalent to 1.602 times 10 to the power of minus 19 joules. A megaelectron volt is 10 to the power of six times larger than an electron volt. So the number of joules in a megaelectron volt will also be 10 to the power of six times larger than the number of joules in an electron volt. 1.602 times 10 to the power of minus 19 joules times 10 to the power of six is 1.602 times 10 to the power of minus 13 joules.

As we want to convert to units of megaelectron volts, we need to multiply the binding energy in joules by one megaelectron volt. And we divide one megaelectron volt by 1.602 times 10 to the power of minus 13 joules. Because they are equivalent, the joule units cancel. And if we perform the calculation, we get a value of 39.28977528 megaelectron volts.

So we have now calculated the binding energy in units of megaelectron volts. But the question asks for the binding energy per nucleon. We have calculated the binding energy for all of the nucleons in the nucleus. So we need to divide this value by the number of nucleons. The question tells us that we have lithium-7. So there are a total of seven protons and neutrons or seven nucleons in the nucleus. So we need to divide 39.28 megaelectron volts by seven. This gives a value of 5.61282504 megaelectron volts. We have now calculated the average binding energy per nucleon in units of megaelectron volts. Note that we call it the average binding energy because the binding energy will vary between protons and neutrons. But weโve just grouped the protons and neutrons and calculated the binding energy per nucleon.

The question also asks for the answer to be given to two decimal places. 5.61282504 megaelectron volts to two decimal places is 5.61 megaelectron volts. So the answer to the question is 5.61 megaelectron volts.

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