Can High Energy Gamma Rays Interact with Protium and Heat It Up?

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Hi All,

How can a high energy (above 1 MeV) gamma ray interact with a protium (hydrogen with only a proton in nucleus)? Does it only ionize it? Does it break it up? Does a chain reaction occur where antiparticles are formed and then annihilated, spitting out another gamma ray? I've been trying to find an answer, and so far haven't found anything. I would greatly appreciate getting a clear answer from anybody.
 
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Gamma rays above 1.022 Mev may break up into a positron-electron pair in the presence of a nucleus, where the nucleus is needed to insure conservation of momentum, but is otherwise unaffected. For a proton the reaction is highly improbable, but the probability does increase with increasing energy of the gamma ray.
 
Does the atom with the nearby nucleus (the one the gamma ray interacts with) heat up during pair production?

Can gamma ray energy be turned into heat? If so, how? I have read some about attenuation, but I still don't fully understand if/how gamma rays can cause something to heat up.
 
I don't know what you have in mind by "heat up". The effect on the nucleus is to add some momentum, i.e. make it move.

Gamma rays energy is converted to heat by a series of steps. There are three main reactions for the first step in gamma ray interaction with matter. These are (1) pair production; (2) scattering off an electron, where some of the gamma ray energy is converted to electron momentum; and (3) the photoelectric effect, where all the energy is used to knock an electron out of an atom. As a consequence of any of these interactions, the energy of the gamma ray is partly or completely transferred to charged particles, which lose energy in reaction with matter. Ultimately this ends up as heat.
 
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I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
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