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kotykmax [81]
3 years ago
7

. Through what three mechanisms do gamma rays interact with matter?

Chemistry
1 answer:
siniylev [52]3 years ago
3 0

Answer:

Photoelectric effect, pair production and Compton scattering

Explanation:

Gamma rays, having no charge, can be slowed slowly by ionization as a material passes through. They suffer other mechanisms that eventually make them disappear, transferring their energy, they can cross several centimeters of a solid, or hundreds of meters of air, without undergoing any process or affecting the material they cross. Then they suffer one of the three effects and deposit much of their energy there. The three mechanisms of interaction with matter are: the photoelectric effect, the Compton effect and the production of pairs.

The photoelectric effect is that the photon meets an electron in the material and transfers all its energy, disappearing the original photon. The secondary electron acquires all the energy of the photon in the form of kinetic energy, and is sufficient to separate it from its atom and convert it into a projectile. This is stopped by ionization and excitation of the material

In the Compton effect the photon collides with an electron as if it were a clash between two elastic spheres. The secondary electron acquires only part of the energy of the photon and the rest takes it with another photon of lesser energy and diverted.

When an energy photon approaches the intense electric field of a nucleus, the production of pairs can happen. In this case the photon is transformed into an electron positron pair. Since the sum of the mass of the pair is 1.02 MeV, it cannot happen if the photon's energy is less than this amount. If the energy of the original photon is greater than 1.02 MeV, the surplus is distributed by the electron and the positron as kinetic energy, and the material can be ionized. The positron at the end of its path forms a positronium and then annihilates producing two annihilation photons, 0.51 MeV each.

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Rudiy27
The equation for the first dissociation is:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
The acid dissociation constant, Ka1 is 9.0 x 10⁻⁸ 
Construct ICE table and obtain their equilibrium concentrations:
                  H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
I (M):             0.1                            0              0
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9.0 x 10⁻⁸ = \frac{X^{2} }{0.1-x}
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[H₃O⁺] = x = 9.4 X 10⁻⁵ M
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The equation for the second dissociation of the acid is:
HS⁻(aq) + H₂O(l) ⇄ S⁻²(aq) + H₃O⁺(aq)
The acid dissociation constant Ka2 is 1.0 x 10⁻¹⁷
Construct ICE table and obtain their equilibrium concentrations:
                    HS⁻(aq) + H₂O(l) ⇄ S²⁻(aq) + H₃O⁺(aq)
I (M):          9.4 x 10⁻⁵                   0              9.4 x 10⁻⁵
C (M):            -x                            +x            +x
E (M):     9.4 x 10⁻⁵ -x                    x              9.4 x 10⁻⁵ + x
So:
1.0 x 10⁻¹⁷ = \frac{(x)(9.4 x 10^{-5} + x) }{(9.4 x 10^{-5} - x) }
x = 1.0 x 10⁻¹⁷ M
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[S²⁻] = x = 1.0 x 10⁻¹⁷ M
[H₃O⁺] = 9.4 x 10⁻⁵ + 1.0 x 10⁻¹⁷ M = 9.4 x 10⁻⁵ M
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