An Er: YAG laser beam of spot A = 1 mm^2 impinges on tissue. Assume that tissue
ID: 1620293 • Letter: A
Question
An Er: YAG laser beam of spot A = 1 mm^2 impinges on tissue. Assume that tissue is mostly holding water. The absorption coefficient for this laser is alpha = 1000 cm^-1. Calculate. a) The energy required to raise one (1) gram of tissue from body temperature to 45 degree C. Assume that all the energy delivered by the laser is converted into heat and the specific heat c to be equal to 1 cal/g.deg. b) The energy required in Joule (J) to vaporize one (1) gram of tissue from body temperature (H = 540 cal/g). c) How long it takes to vaporize a disk of tissue of thickness L if the laser is running continuously and the power is P = 10W. Make sure you read the comment on the absorption coefficient on page 4 in Unit Two in your course manual before you attempt this question. You may want to read again section 1.5 as it also pertains to tissue absorption. d) The mass of the disk of tissue of thickness L in part c). Assume that tissue is only holding water. Make sure you understand EExplanation / Answer
part a:
mass=1 gram
specific heat=1 cal/(g.deg)
body temperature=37 degree celcius
energy required=mass*specific heat*(final temperature -initial temperature)
=1*1*(45-37)=8 cal.
part b:
energy required to vaporize the tissue =
energy required to increase temperature from 37 degree celcius to 100 degree celcius + energy required to vaporize it
=mass*specific heat*(100-37)+mass*H
=1*1*63+1*540 cal
=603 cal
=2522.952 Joules
part c:
energy required=2522.952 Joules
let time=t seconds
output power/input power=absoprtion coefficient*thickness=1000*L
output power=input power*absorption coefficient
==>2522.952/t=10*1000*L
==>t=0.2523/L seconds
part d:
area=1 mm^2=0.01 cm^2
volume=area*thickness=0.01*L cm^3
mass=volume*density
=0.01*L cm^3*(1 gram/cm^3)
=0.01*L grams
part e:
time period=0.2523/L seconds
laser is active for 0.05 ms per every 0.1 ms
so total time taken=(.2523/L)*0.1/(0.05)=0.5046/L seconds