During laser vision correction, a brief burst of 193-nm ultraviolet light is projected onto the cornea of a patient. It makes a spot 0.80 mm in diameter and evaporates a layer of cornea 0.30μm0.30μm size 12{0 "." "30" mm} {} thick. Calculate the energy absorbed, assuming the corneal tissue has the same properties as water; it is initially at 34ºC34ºC size 12{"34"°C} {}. Assume the evaporated tissue leaves at a temperature of 100ºC100ºC size 12{"100"°C} {}.

*Strategy*

The energy from the laser light goes toward raising the temperature of the tissue and also toward evaporating it. Thus we have two amounts of heat to add together. Also, we need to find the mass of corneal tissue involved.

*Solution*

To figure out the heat required to raise the temperature of the tissue to 100ºC100ºC size 12{"100"°C} {}, we can apply concepts of thermal energy. We know that

Q = mcΔT,Q = mcΔT, size 12{ ital "Q "= ital " mc"ΔT} {}

(6)where Q is the heat required to raise the temperature, ΔTΔT size 12{DT} {} is the desired change in temperature, mm size 12{m} {} is the mass of tissue to be heated, and cc size 12{c} {} is the specific heat of water equal to 4186 J/kg/K.

Without knowing the mass
m
m
at this point, we have

Q=m(4186 J/kg/K)(100ºC–34ºC)=m(276,276 J/kg)=m(276 kJ/kg).Q=m(4186 J/kg/K)(100ºC–34ºC)=m(276,276 J/kg)=m(276 kJ/kg). size 12{Q=m \( "4186 J/kg/K" \) \( "100"°C - "34"°C \) =m \( "276,276 J/kg" \) =m \( "276 kJ" \) } {}

(7)The latent heat of vaporization of water is 2256 kJ/kg, so that the energy needed to evaporate mass mm size 12{m} {} is

Qv=mLv=m(2256 kJ/kg).Qv=mLv=m(2256 kJ/kg). size 12{Q rSub { size 8{v} } = mL rSub { size 8{v} } = m \( "2256" \) " kJ"} {}

(8)To find the mass mm size 12{m} {}, we use the equation ρ=m/Vρ=m/V size 12{ρ= {m} slash {V} } {}, where ρρ size 12{ρ} {} is the density of the tissue and VV size 12{V} {} is its volume. For this case,

m
=
ρ
V
=
(1000 kg/m
3
)
(
area
×
thickness
(m3))
=
(1000 kg/
m3
)
(
π
(
0.80
×
10
–
3
m
)
2
/
4
)
(
0
.
30
×
10
–
6
m
)
=
0.151
×
10
–
9
kg.
m
=
ρ
V
=
(1000 kg/m
3
)
(
area
×
thickness
(m3))
=
(1000 kg/
m3
)
(
π
(
0.80
×
10
–
3
m
)
2
/
4
)
(
0
.
30
×
10
–
6
m
)
=
0.151
×
10
–
9
kg.
alignl { stack {
size 12{m = ρ"V "=" 1000 kg/m" rSup { size 8{3} } \( "area"´"thickness" \) } {} #
="1000" \( p \( 0 "." "80"´"10" rSup { size 8{ +- 3} } " m" \) rSup { size 8{2} } /4 \) \( 0 "." "30"´"10" rSup { size 8{-6} } " m" \) {} #
=0 "." "151"´"10" rSup { size 8{ +- 9} } " kg" "." {}
} } {}

(9)Therefore, the total energy absorbed by the tissue in the eye is the sum of QQ size 12{Q} {} and QvQv size 12{Q rSub { size 8{v} } } {}:

Qtot=m(cΔT + Lv)=
(0.151×10−9kg)
(276 kJ/kg+2256 kJ/kg)=382×10−9 kJ.Qtot=m(cΔT + Lv)=
(0.151×10−9kg)
(276 kJ/kg+2256 kJ/kg)=382×10−9 kJ. size 12{Q rSub { size 8{"tot"} } = m \( cD"T "+" L" rSub { size 8{v} } \) =0 "." "151"´"10" rSup { size 8{-9} } \( "276"+"2256" \) ="382"´"10" rSup { size 8{-9} } " kJ"} {}

(10)
*Discussion*

The lasers used for this eye surgery are excimer lasers, whose light is well absorbed by biological tissue. They evaporate rather than burn the tissue, and can be used for precision work. Most lasers used for this type of eye surgery have an average power rating of about one watt. For our example, if we assume that each laser burst from this pulsed laser lasts for 10 ns, and there are 400 bursts per second, then the average power is Qtot×400=150 mWQtot×400=150 mW size 12{Q rSub { size 8{"tot"} } ´"400"="150"" mW"} {}.

Comments:"This introductory, algebra-based, two-semester college physics book is grounded with real-world examples, illustrations, and explanations to help students grasp key, fundamental physics concepts. […]"