Compare and contrast both densely-ionizing and sparsely-ionizing radiation.

Physics

1 answer:

kherson [118]3 years ago

7 0

Explanation:

Radiation biology is the study of the effects of ionizing radiation on living organisms, when this radiation is absorbed in biologic material, ionization and excitations occur in a pattern that depends on the type of radiation involved, considering how far the primary ionization events are separated in space, radiation can be sparsely ionizing (for example, an x-ray or a gamma-ray) or densely ionizing (for example, a proton or an alpha particles).

<em>DNA is the most critical biological target because of the limited redundancy of the genetic information it contains, most lesions are reparable, but those produced by densely ionizing radiation are generally less reparable than those produced by sparsely ionizing radiation. Densely ionizing (high LET) radiations, therefore, typically have a higher relative biological effectiveness.</em>

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Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57

Harlamova29_29 [7]

We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
$n_1sin(\theta_1)=n_2sin(\theta_2)$
Where $\theta_2$ differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, $\Delta x$ is the distance between both rays.
$\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705$
$\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21$
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
$d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m$
For red we have:
$d_{red}=h.tan(\theta_{2red})\approx 0.0134m$
We finally have:
$\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m$