Photon+deuteron->p+n , E(photon)=?

  • Thread starter Thread starter Ene Dene
  • Start date Start date
Ene Dene
Messages
47
Reaction score
0
[SOLVED] photon+deuteron-->p+n , E(photon)=?

Problem:
What is the minimal energy of photon for disassembling deuteron on proton and neutron (photon+d --> p+n)? By how much is that energy larger than energy of binding of deuteron? Binding energy of deuteron is E=2.225MeV, mass of deuteron is 1875.628MeV.


This means that m(d)-m(p)-m(n)=2.225MeV.
I don't understand why would do you need a photon of greater energy than 2.225MeV if that is the energy that binds proton and neutron together.
 
Physics news on Phys.org
Ene Dene said:
Problem:
What is the minimal energy of photon for disassembling deuteron on proton and neutron (photon+d --> p+n)? By how much is that energy larger than energy of binding of deuteron? Binding energy of deuteron is E=2.225MeV, mass of deuteron is 1875.628MeV.


This means that m(d)-m(p)-m(n)=2.225MeV.
I don't understand why would do you need a photon of greater energy than 2.225MeV if that is the energy that binds proton and neutron together.
Binding energy is not energy that binds nucleons, but rather it is the energy given off when a nucleon combines with another nucleon or nucleus, i.e. binding energy is actually the energy require to unbind a nucleon from the nucleus.

In the case of the deuteron, the energy required to unbind the neutron and proton is 2.225 MeV. That would be approximately the energy of a gamma-ray given off when a proton and neutron combine (neutron capture) to form a deuteron.
 
Astronuc said:
Binding energy is not energy that binds nucleons, but rather it is the energy given off when a nucleon combines with another nucleon or nucleus, i.e. binding energy is actually the energy require to unbind a nucleon from the nucleus.

In the case of the deuteron, the energy required to unbind the neutron and proton is 2.225 MeV. That would be approximately the energy of a gamma-ray given off when a proton and neutron combine (neutron capture) to form a deuteron.

But if Iunderstand correctly the OP, th equestion was ho wmuch above the binding energy must the photon have. And the OP was wondering about why some energy above the binding energy was required.

The answer is of course conservation of 4-momentum. Some of the photon energy goes into breaking apart the deuteron but some of it is needed to give some kinetic energy to the neutron and proton.
 
For a photon of energy p (with c=1),
p+M_d=\sqrt{(m_p+m_n)^2+p^2}.
Square and solve for p.
 
Astronuc said:
Binding energy is not energy that binds nucleons, but rather it is the energy given off when a nucleon combines with another nucleon or nucleus, i.e. binding energy is actually the energy require to unbind a nucleon from the nucleus.
Yes, I did think that these energies were the same, that's why I asked a question, why aren't they the same.
kdv said:
But if understand correctly the OP, th equestion was ho wmuch above the binding energy must the photon have. And the OP was wondering about why some energy above the binding energy was required.
Yes, you did understand correctly but I also wanted to know why is that so.
And here is why:
kdv said:
The answer is of course conservation of 4-momentum. Some of the photon energy goes into breaking apart the deuteron but some of it is needed to give some kinetic energy to the neutron and proton.
pam said:
For a photon of energy p (with c=1),
p+M_d=\sqrt{(m_p+m_n)^2+p^2}
Square and solve for p.

Thank you all, now I understand.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top