Answer:
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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Answer and Explanation:
- Should let force applied throughout horizontal direction become "F", as well as the displacement of that same sled, become "d." This same force's work seems to be "Fd."
- Presently, unless the mass would be doubled as well as the force is applied continues to remain the very same, the work used by the force to keep moving the sled by such the distance "d" would then remain the same.
So that the above seems to be the right answer.
Wavelength i’m assuming that would be the most reasonable
