Question

a photon beam is made from a target with a high impedance. electron beams of 6 and 16Mev are used in the process, shot at the target. the ratio of beam intensity is measured in the direction of the electron beam (0 degrees) and at 25 degrees from the angle of incidence.nFor what electron bean energy (and why) is the ratio of intensities (i₀/i₂₅) the largest?

Answer 1

I am not quite sure the applicability of the formula which I am going to use since the paper which I have read about the bremsstrahlung(what is happening in this situation) suggests only the numerical estimation of the angular dependence of the radiation.

The angular dependence of the emitted radiation in bremsstrahlung is given by

                       

where

q is the charge of the particle.

a is the acceleration of the particle.

is the angle between the acceleration and the direction of observation.

is given by v/c where v is the velocity of the particle.

In our case, we have to take the ratio of the intensity emitted at the angle of o and 25 degrees respectively. Although the sin of zero is zero and in that case, the ratio will turn out to be equal but we assume that we are looking at very close to the direction of the electron beam than we will be having the same terms independent of the energy of the particle. Means, that term will not affect the answer and we can omit that. So finally the formula gets simplified and is given by

where has to be determined at the energy of the particle which in our case turns out to be

that ratio for energy 6 MeV and 16 MeV turns out to be 45 and 180 respectively.

Hence, the ratio of the intensity of the emitted radiation at 0 and 25 degrees is more when the energy of the incident electron is high.

Summary:

1) We have determined this ratio in the limiting case when we are looking at the radiation very closely.

2) This is the situation of bremsstrahlung.