Video Transcript
Two isolated protons and two
isolated neutrons have a cumulative mass of 4.03188 unified atomic mass units. A helium-4 nucleus has a mass of
4.0015 unified atomic mass units. What is the binding energy per
nucleon for the helium-4 nucleus? Your answer should have the
mega-electron volt unit and be rounded to one decimal place.
To answer this question, we need to
calculate the binding energy for the helium-4 nucleus. Binding energy is the minimum
energy required to disassemble an atomic nucleus into unbound protons and
neutrons. A helium-4 nucleus consists of two
protons and two neutrons. If the binding energy was added to
the helium-4 nucleus, the nucleus would separate into two isolated protons and two
isolated neutrons.
The question gives us the mass of a
helium-4 nucleus and the cumulative mass of two isolated protons and two isolated
neutrons. The difference between these masses
is known as the mass defect. We can calculate the mass defect by
subtracting the mass of the helium-4 nucleus from the mass of the isolated
nucleons. This gives us a mass defect of
0.03038 unified atomic mass units.
The mass defect and binding energy
can be related using Einstein’s equation. Einstein’s equation can be
represented as Δ𝐸 equals Δ𝑚𝑐 squared. For the purposes of this question,
Δ𝐸 represents the binding energy in units of joules. Δ𝑚 is the mass defect in
kilograms. 𝑐 is the speed of light, a
constant that can be approximated to three times 10 to the eighth meters per
second.
Before we can substitute the mass
defect into the equation and solve for the binding energy, we first need to convert
the mass in unified atomic mass units to kilograms. To do this, we’ll need to use the
relationship one unified atomic mass unit equals 1.66 times 10 to the negative 27th
kilograms. To perform the conversion, we
multiply the mass defect in unified atomic mass units by the conversion factor
written as a fraction with unified atomic mass units in the denominator so that the
units cancel. Performing the calculation gives us
a mass defect of 5.04308 times 10 to the negative 29th kilograms.
Now, we can substitute the mass
defect in kilograms and the speed of light in meters per second into the binding
energy equation. Performing the calculation gives us
a binding energy of 4.53877 times 10 to the negative 12th kilogram meters squared
per second squared. The unit kilogram meters squared
per second squared is the same as the unit joule.
We have calculated the binding
energy. But we are told in the question
that our answer should have the unit mega-electron volts. To convert the binding energy from
joules to mega-electron volts, we’ll need to know two conversion factors: one
electron volt is equal to 1.60 times 10 to the negative 19th joules and one
mega-electron volt is equal to one times 10 to the sixth electron volts. We can start by converting the
binding energy in joules to electron volts by multiplying by one electron volt per
1.60 times 10 to the negative 19th joules. Performing the calculation, the
joule units will cancel, giving us 28367325 electron volts.
Now we can convert the energy into
mega-electron volts by multiplying the energy in electron volts by one mega-electron
volt per one times 10 to the sixth electron volts. Performing the calculation gives us
a binding energy of 28.3673 mega-electron volts.
Our answer is now in the correct
unit, but the question asked for the binding energy per nucleon. A helium-4 nucleus has four
nucleons. So we simply need to divide the
binding energy by four. This gives us a binding energy per
nucleon of 7.09183 mega-electron volts. Lastly, we need to round our answer
to one decimal place. Rounding appropriately, we have
determined that the binding energy per nucleon for the helium-4 nucleus is 7.1
mega-electron volts.
