Quantum Physics in the Body: Energy

AI Thread Summary
The discussion centers on calculating the energy, power output, and photon delivery of a pulsed dye laser used in medical procedures. The energy required to vaporize 2.5 µg of blood is determined to be approximately 6.34 mJ, factoring in both the heat of vaporization and the temperature increase needed to reach the boiling point. The power output of the laser is calculated to be around 13.91 W based on the energy delivered per pulse duration of 456 µs. The participants successfully navigated through the calculations, confirming the accuracy of their results. Overall, the thread highlights the application of quantum physics principles in medical laser technology.
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Homework Statement



A pulsed dye laser emits light of wavelength 576 nm in pulses of 456 µs duration. This light is absorbed by the hemoglobin in the blood and can therefore be used to remove vascular lesions, such as certain blemishes and birthmarks. To get an estimate for the power required for this laser surgery, assume that blood has the same specific heat capacity (4190 J/(kg·K)) and heat of vaporisation (2.256×106 J/kg) as water. Suppose that each pulse must remove 2.5 µg of blood by evaporating it, starting at 33 °C.
a) How much energy must the pulse deliver to the blemish?
b) What must be the power output of the laser?
c) How many photons does each pulse deliver?

Homework Equations



Q = m*c*delta(t)
f = c/wavelength
E = h*f

The Attempt at a Solution



I don't even know how to start this... but I'd really appreciate the help, especially an answer to part a.
 
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Okay... I'm starting to make some progress, but I keep getting stuck in my ideas...

Energy to Vaporize = Q = m*l = (2.5e-9 kg)(2.256×106 J/kg) = 5.64e-03 J...
so...
Q = m*c*delta(t) ; delta(t) = Q/(m*c) = 5.64e-03 J/[(2.5e-9 kg)x(4190 J/(kg·K))]
delta(t) = 538.42 K

but... I don't understand... This means there's a difference...
33°C = 306.15 K, so there's a difference of 232.275 K.

What can I do with this?
 
Your thermal energy for vaporizing blood is correct. But you start out at 33°C and blood might not undergo a phase transition then, right? You might have to add additional heat to get the temperature high enough to reach the temperature at which blood transitions from a liquid to a gas.

It looks like you attempted to find the temperature at which blood transitions from a liquid to a gas. My guess is is that you're supposed to use the boiling point of water.
 
Ah, that's very interesting!
I never though of using that 100°C boiling point.
so, with that...
Energy to Vaporize = Q = m*l = (2.5e-9 kg)(2.256×106 J/kg) = 5.64e-03 J
and
Q = m*c*delta(t) = (2.5e-9 kg)(4190 J/(kg·K))(67K) = 7.01825e-4 J

adding these two values... Q1 +Q2 = 6.341825e-3 J, which makes sense...

now,
b) What must be the power output of the laser?
c) How many photons does each pulse deliver?

ideas there?
 
Well how do we define power?
 
Okay, I figured out b...

I thought, since a J/s is a W...

Then the energy, (6.341825e-3 J)/0.000456s = 13.9076 W...

so far, so good... now, c.
 
I figured out c too!
Thank you so much for your help!
 
Looks right so far.

EDIT: Nice, good job.
 
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