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

minimum number of photon is 4.05 ×
`10^(7)`

Step-by-step 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)`