3
$\begingroup$

I've been reading on radioactivity but along the way I got confused, if binding energy is the amount of energy used in holding the nucleus together then why is binding energy also the amount of energy used in breaking the nucleus apart, shouldn't the energy required to break the nucleus apart be greater than the initial energy?

$\endgroup$
  • $\begingroup$ Binding energy of a system is the energy required to break that system apart. Equivalently, it's the height of the potential barrier that is holding the system together. $\endgroup$ – user1247 Apr 7 at 21:40
  • $\begingroup$ Read your statement again. "binding energy IS the amount of energy used in holding the nucleus together". If you provide that much energy, constituents of the nucleus don't stay 'bound'. Where are you getting confused? $\endgroup$ – Avantgarde Apr 7 at 21:43
  • $\begingroup$ @Avantgarde: if a force holds the nucleons together, then the force to counter this force must be greater so as to break the bond $\endgroup$ – Tak Apr 7 at 21:45
  • 1
    $\begingroup$ @Tak The binding energy is defined as the energy you need to supply that exactly counters the binding forces, with the final products having zero kinetic energy. If you provide energy more than the binding energy, you will not only break the nucleus apart, but also impart additional energy to the final products in the form of kinetic energy. $\endgroup$ – Avantgarde Apr 7 at 21:47
  • $\begingroup$ @Avantgarde: but the binding energy in fusion reactions is released, so would there be any force holding the nucleons, since the binding energy has been released? $\endgroup$ – Tak Apr 7 at 21:53
0
$\begingroup$

This is very basic: we are looking for the minimum energy supply that we must apply.

This is very similar to that thing students use to ask: "why do we need a force $mg$ to overcome weight? It should be greater". And, okay, rigurously speaking, if youapply a force $mg$ upwards, you won't rise the weight. But you will have cancelled the weight force, so the object will now behave like a free body.

So, if you apply the tiniest increment, say $10^{-9}$ more newtons, for example, that will be enough to rise it, because now the body is free. But I could have also chosen $10^{-12}N$, and the object would also go upwards. The acceleration would be really small, true, but that wuold be enough to rise it.

So, we obviously need to give a number (not any than I can think of, but an objective figure). We choose to say that you must apply $mg$ to cancel weight, and then, the more you apply, the more you'll get. It is understood that it will have to be greater, but how greater is up to you. We just give the lower limit: the one that jsut cancels it out.

So this is the same. If we choose a free body to have $0 J$ wenergy, then any bound state has less than $0J$, (negative energy), because it hasn't got enough energy to reach freedom, so it must be less than $0J$.

Let's call that energy $E_b$, and $E_B<0$, so the body has energy $-|E_b|$. You must apply $+E_b$ to overcome that. Like that, the particle would be able to reach freedom (probably at infinity), without any KE. If you want some extra $KE$, that's the extra energy you need to supply.

$\endgroup$
-2
$\begingroup$

You are on the right path, but it needs a little explanation. Let's go from the smallest known particles up.

Let's say that the neutron is made of 3 quarks (it's actually made of a sea of quarks, antiquarks, and gluons, antigluons, the net is called valence quarks), those quarks do have rest mass, right? Sure. Add them up, and you get 1% of the rest mass of the neutron.

Where does the 99% of the rest mass of the neutron come from? It is the binding energy between the quarks, called color force.

Now let's go one size up. Lets look at the nucleus. It is made of neutrons and protons. Same way. The rest mass of the neutrons and protons will add up to 1% of the rest mass of the nucleus. Where does the 99% of the rest mass of the nucleus come from? Binding energy between the neutrons and protons called residual strong force, or nuclear force.

From wiki:

The total rest masses of the fission products (Mp) from a single reaction is less than the mass of the original fuel nucleus (M).

If you want to tear a nucleus apart, and that is nuclear fission, you need to add at least as much energy as the energy that binds the constituents together. When you reach that energy limit, you will get a peanut nucleus. What the energy limit is, depends on how much energy binds the nucleus's constituents (neutrons and protons) together. That energy is the residual strong force or nuclear force. When you add enough energy (meaning you add a high energy neutron), that is at least as much as the binding energy, the nucleus will fission into two separate daughter nuclei. First, it will be a peanut nucleus, and at that point, the EM repulsion between protons in the daughter nuclei will be enough (it will overcome the nuclear force) to shoot the two daughter nuclei apart at 3% the speed of light.

$\endgroup$
  • 1
    $\begingroup$ Are you sure that “The rest mass of the neutrons and protons will add up to 1% of the rest mass of the nucleus”? $\endgroup$ – HolgerFiedler Apr 8 at 5:30
  • $\begingroup$ This answer has a lot of issues. The relativistic QCD vacuum, whose properties mean that the "bare quark masses" sum to much less than the mass of a nucleon, isn't a good example of a binding energy. The mass of a typical nucleus is $\sim 99\%$ of the mass of its constituent nucleons, as is usually verified (by doing arithmetic) during introductory discussions about why fusion and fission release energy. This answer says that the bare nucleons are much less massive than the bound nucleus, which is not correct. Also I know of no fission model that includes a "donut nucleus." $\endgroup$ – rob Apr 8 at 17:06
  • $\begingroup$ @rob wow. sorry, it is peanut, not doughnut. thank you I edited. And about the mass, this is from wiki "The total rest masses of the fission products (Mp) from a single reaction is less than the mass of the original fuel nucleus (M). " $\endgroup$ – Árpád Szendrei Apr 8 at 17:44
  • 1
    $\begingroup$ @ÁrpádSzendrei You are misinterpreting that sentence from Wikipedia. It's correct for fission of very heavy nuclei into medium-mass nuclei, but it's not correct for removing a single nucleon from any (non-dripline) nucleus, and it's not correct for disassembling a nucleus completely into nucleons (which is how the binding energy is properly defined). $\endgroup$ – rob Apr 8 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.