Question

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%)

Answer 1

Answer:

minimum number of photon is 4.05 × $10^{7}$

Explanation:

given data

energy = 50 keV = 50 × $10^{3}$ eV =  50 × $10^{3}$ × 1.602× $10^{-19}$  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 × $10^{3}$ = n × 6.6× $10^{-34}$ × c/ 1× $10^{-3}$

50 × $10^{3}$ × 1.602× $10^{-19}$  = n × 6.6× $10^{-34}$ ×( 3 × $10^{8}$ / 1× $10^{-3}$)

solve it and find n

n = 4.05 × $10^{7}$

so here minimum number of photon is 4.05 × $10^{7}$