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

  • Thread starter Ene Dene
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In summary, the minimal energy of a photon for disassembling a deuteron into a proton and neutron is given by the sum of the masses of the proton and neutron plus the binding energy of the deuteron, and this is due to conservation of 4-momentum. This energy must be greater than the binding energy of the deuteron as some of the photon's energy goes into breaking apart the deuteron and giving kinetic energy to the proton and neutron. This is why the photon must have energy greater than the binding energy of the deuteron.
  • #1
Ene Dene
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[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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
For a photon of energy p (with c=1),
[tex]p+M_d=\sqrt{(m_p+m_n)^2+p^2}[/tex].
Square and solve for p.
 
  • #5
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),
[tex]p+M_d=\sqrt{(m_p+m_n)^2+p^2}[/tex]
Square and solve for p.

Thank you all, now I understand.
 

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

What is the process of "Photon+deuteron->p+n"?

The process of "Photon+deuteron->p+n" is a nuclear reaction in which a photon and a deuteron (a nucleus of deuterium, consisting of one proton and one neutron) collide and produce a proton and a neutron as products.

What is the energy of the photon in this process?

The energy of the photon in this process is denoted by E(photon) and can vary. It is typically measured in units of electron volts (eV) or joules (J).

What is the purpose of studying this reaction?

The purpose of studying this reaction is to better understand the interactions between particles and to gain insights into the structure and properties of nuclei.

Can this reaction occur at any energy level?

Yes, this reaction can occur at any energy level. However, the probability of the reaction happening increases as the energy of the photon increases.

What other particles can be produced in this reaction?

In addition to a proton and a neutron, other particles that can be produced in this reaction include pions, kaons, and hyperons. These particles are formed when the energy of the photon is high enough to create more massive particles.

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