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 ×
Answer:
3234.2 W
Explanation:
Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.
So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².
Now, the power radiated on the patch of area 0.570 m² at the equator is
P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W
Answer:
q = 8472.2 W/m
Explanation:
The solving or solution is given in the attach documents.
The answer for this question is: C
