(a) During laser vision correction, a brief burst of 193 nm ultraviolet light is
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Question
(a) During laser vision correction, a brief burst of 193 nm ultraviolet light is projected onto the cornea of the patient. It makes a spot 0.800 mm in diameter and evaporates a layer of cornea 0.300 µm thick. Calculate the energy absorbed assuming the corneal tissue has the same properties as water and is initially at 34.0 oC. The evaporated tissue leaves at a temperature of 100 oC. (b) If each laser burst lasts 10.0 ns, what is the power of the burst? Assume the energy found in (a) is the entire energy. (c) If there are 10 bursts per second, what is the average power output?
Explanation / Answer
(a)
To figure out the heat required to raise the temperature of the tissue to 100ºC , we can apply concepts of thermal energy.
We know that Q = mcT
where Q is the heat required to raise the temperature, T is the desired change in temperature, m is the mass of tissue to be heated, and c is the specific heat of water equal to 4186 J/kg/K.
Without knowing the mass m at this point,
we have Q = m(4186 J/kg/K)(100ºC – 34ºC) = m(276,276 J/kg) = m(276 kJ/kg).
The latent heat of vaporization of water is 2256 kJ/kg, so that the energy needed to evaporate mass m is
Qv = mLv = m(2256 kJ/kg)
To find the mass m , we use the equation = m / V , where is the density of the tissue and V is its volume. For this case,
m = V
= (1000 kg/m3 )(area×thickness(m3 ))
= (1000 kg/m3 )((0.80×10 – 3 m)2 / 4)(0.30×10 – 6 m)
= 0.151×10 – 9 kg
Therefore, the total energy absorbed by the tissue in the eye is the sum of Q and Qv :
Qtot = m(cT + Lv) = (0.151×109 kg)(276 kJ/kg + 2256 kJ/kg) = 382×109 kJ..........Ans.
(b) Power = Energy/time = 382×109 kJ /10.0x10^-9 s = 38.2 Watt...........Ans.
(c) If 10 burst per seconds , then Average Power = 382 x 10^-9 J /10 x 10^-8 = 382 watt
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