Question

Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll. ratio. (b) Plexiglas is often used to shield high-energy beta emitters rather than lead, even though lead is a better shield against the bremsstrahlung photons. Both shields will stop the high-energy beta, so why is Plexiglas used instead of lead?

Answer 1

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

$Q = \dfrac{4mME }{(m+M)^2}$

However; from the total stopping power & power loss of the electron;

$\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}$

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss  $=\dfrac{82 \times 1.9}{800}$

= 0.19475

b)

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.