Calculate minimum number of incident photons per area and of the minimum dose needed to visualize an object of 1 mm squared using x rays of energy 50 kev. ( thickness 10^-3)(contrast 1%)

minimum number of photon is 4.05 ×

Explanation:

given data

energy = 50 keV = 50 × eV =  50 × × 1.602×  J

thickness = 10^-3

contrast = 1%

to find out

number of incident photons

solution

we know here equation that is

E  = n  × h  × ν   .......................1

put here all these value

50 × = n × 6.6× × c/ 1×

50 × × 1.602×  = n × 6.6× ×( 3 × / 1× )

solve it and find n

n = 4.05 ×

so here minimum number of photon is 4.05 ×

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