91Ó°ÊÓ

The wavelength of the laser light$\mathrm{λ}=585\mathrm{nm}=585×{10}^{-9}\mathrm{m}$

Pulse duration$\mathrm{t}=450\mathrm{μ²õ}=450×{10}^{-6}\mathrm{s}$

The specific heat$\mathrm{c}=4190{\mathrm{JKg}}^{-1}{\mathrm{K}}^{-1}$

The heat of vaporization as water${\mathrm{c}}_{\mathrm{vap}}=2.056×{10}^6{\mathrm{JKg}}^{-1}$

The amount of blood removed in each pulse$\mathrm{m}=2\mathrm{μ²\mu }=2×{10}^{-9}\mathrm{Kg}$

Starting temperature ${\mathrm{T}}_1={33}^{\mathrm{ο}}\mathrm{C}={306}^{\mathrm{ο}}\mathrm{K}$

Final or vaporization temperature of the water ${\mathrm{T}}_2={100}^{\mathrm{ο}}\mathrm{C}={373}^{\mathrm{ο}}\mathrm{K}$

Now as we know that the energy must each pulse deliver to the blemish is given by

$$\begin{gathered} Q=mcâ–³T+m{c}_{vap} \\ Q=mc(T_2-T_1)+m{c}_{vap} \end{gathered}$$ (1)

Where m is mass, c is specific heat, andis the change in temperature

Hence,

$$\begin{gathered} \mathrm{Q}=2×{10}^{-9}\mathrm{kg}×[4190{\mathrm{Jkg}}^{-1}{\mathrm{K}}^{-1}×({373}^{∘}\mathrm{K}-{306}^{∘}\mathrm{K})+2.256×{10}^6{\mathrm{Jkg}}^{-1}] \\ =5.1×{10}^{-3}\mathrm{J} \\ =5.1\mathrm{mJ} \end{gathered}$$ (2)

The power is given by

$P=\frac{E}{t}=\frac{Q}{t}$

So from equations (2) and (3), we get the power output of this laser will be

$$\begin{gathered} P=\frac{5.0×{10}^{-3}J}{450×{10}^{-6}s} \\ P=11.33J{s}^{-1} \\ P=11.33W \end{gathered}$$

As we know that the energy of one photon is

$E=\frac{hc}{\mathrm{λ}}$ (4)

Hence the number of photons emitted by each pulse that is delivered to the blemish will be the total energy delivered to the blemish divided by the energy of a single photon i.e.

$$\begin{gathered} N=\frac{Q}{E}=\frac{Q\mathrm{λ}}{hc} \\ N=\frac{5.0×{10}^{-3}-3)J×585×{10}^{-9}m}{6.626×{10}^{-34}J.s×3×{10}^{8}m{s}^{-1}} \\ N=1.5×{10}^{16} \end{gathered}$$