Question

During laser vision correction, a brief burst of 193-nm ultraviolet light is projected onto the cornea of a patient. It makes a spot 0.50 mm in diameter and evaporates a layer of cornea 0.25 um thick. Calculate the energy absorbed, assuming the corneal tissue has the same properties as water; it is initially at 34°C. Assume the evaporated tissue leaves at a temperature of 100°C. Provide your answer in micro-joules (ud). __124 -- UJ. Wrong Answer

Answer 1

solution) - 193 nm - 1938 10-im D=0.50mm = 0.50x 103 m t = 0.25 um = 0.25x15m T-34c = 30 + 273) ok T2 = 100 c = (100 + 273) ok we have, Q = MCAT + mL C = 4186 I leglk 6 kolkg = 2256x10 L-2256 kg 1kg = 22 AT = T - T, = 100+27 3-3 AT= 66k -8 (Axt). m=pv for w map(1984)xt) (8=1000 kg/m3 me (1000) (+ (0-50xlo ") }(0-26418 m= 4.9 1x10 "/kg Q = MCAT & ML = m (CAT + L) Q = (4.91x.- (4186 x 66.+2256x/o?) h to Q = 124.33x10 6 J O MJ Q = 124.3