A beam of electrons of energy4800keV is allowed to fall on a

A beam of electrons of energy4800.keV is allowed to fall on a Gold target Z = 79 to produce x-rays. Calculate the percentage (%) of energy that is lost to (fray and percentage of energy that is utilized to produce x-rays If the current earned by the electron beam in (a) is 0.4 A then find the efficiency of the system to produce x-rays. An x-ray film is exposed to x-rays of intensity 4 x I0^15. After exposure, you found that the Optical Density of the film is 2.8, as measured by a densitometer. Find the intensity of the transmitted x-rays. Find the contrast of the low speed film B in the bellow figure. For Graduates, only: Find the sensitivity of film B in the above figure.

Solution

GIVEN:- E = 4800 kev = 4.8 Mev, Zgold = Z = 79,

SOLUTION:-

(b) By formula efficiency to produce Z-rays,   = 0.0007 x Z x E = 0.0007 x 79 x 4.8 = 0.265 x100 = 26.5%

So efficiency   = 26.5% (ANSWER)

(a) From (b) we observe that = 26.5% i.e. = Output energy / Input energy(E)

i.e. Output energy (O) = E x = 4.8 (Mev) x 0.265 = 1.272 Mev

i.e. % of Energy lost to heat = (4.8 - 1.272) / 4.8 x 100 = 73.5 % (ANSWER)

and % of Energy utilised to produce x-rays = 1.272/4.8 x 100 = 26.5 % (ANSWER)