A pushed dye laser emits light of wavelength 585 nm. Because this wavelength is strongly absorbed by the haemoglobin in the blood, the method is especially effective for removing various types of blemishes due to blood. To get a reasonable estimate of the power required for such laser surgery, we can model the blood as having the same specific heat and heat of vaporization as water. [`S=4.2xx10^(3)J(kgK)^(-1)`,`L=2.25xx10^(6)Jkg`]
Q. Suppose that each pulse must remove `2mug` of blood by evaporating it starting at `30^circ`C. The energy that each pulse must deliver to the blemish is nearly

Options

A. `5.1J`
B. `5.1mJ`
C. `5.1muJ`
D. `5.1kJ`

Explanation

To solve the problem of estimating the energy required for a laser pulse to remove 2 µg of blood by evaporating it starting at 30°C, we will follow these steps:

### Step 1: Convert mass from micrograms to kilograms
The mass of blood to be evaporated is given as 2 µg. We need to convert this mass into kilograms for our calculations.

\[ \text{Mass} = 2 \, \mu g = 2 \times 10^{-6} \, g = 2 \times 10^{-9} \, kg \]

### Step 2: Calculate the energy required to raise the temperature of the blood to its boiling point
We will use the specific heat capacity of water (which we are assuming is the same for blood) to calculate the energy required to raise the temperature of the blood from 30°C to 100°C (the boiling point of water).

The formula for the energy required to change the temperature is:

\[ Q_1 = m \cdot S \cdot \Delta T \]

Where:
- \( m = 2 \times 10^{-9} \, kg \)
- \( S = 4.2 \times 10^{3} \, J/(kg \cdot K) \)
- \( \Delta T = 100°C - 30°C = 70 \, K \)

Now substituting the values:

\[ Q_1 = (2 \times 10^{-9}) \cdot (4.2 \times 10^{3}) \cdot (70) \]

Calculating \( Q_1 \):

\[ Q_1 = 2 \times 10^{-9} \cdot 4.2 \times 10^{3} \cdot 70 = 5.88 \times 10^{-5} \, J \]

### Step 3: Calculate the energy required for vaporization
Next, we calculate the energy required to vaporize the blood once it reaches the boiling point using the latent heat of vaporization.

The formula for the energy required for vaporization is:

\[ Q_2 = m \cdot L \]

Where:
- \( L = 2.25 \times 10^{6} \, J/kg \)

Now substituting the values:

\[ Q_2 = (2 \times 10^{-9}) \cdot (2.25 \times 10^{6}) \]

Calculating \( Q_2 \):

\[ Q_2 = 4.5 \times 10^{-3} \, J \]

### Step 4: Calculate the total energy required
The total energy \( Q \) required for the laser pulse is the sum of the energy required to raise the temperature and the energy required for vaporization:

\[ Q = Q_1 + Q_2 \]

Substituting the values we calculated:

\[ Q = 5.88 \times 10^{-5} + 4.5 \times 10^{-3} \]

Calculating \( Q \):

\[ Q = 0.0000588 + 0.0045 = 0.0045588 \, J \]

### Step 5: Convert to millijoules
To express the energy in millijoules:

\[ Q = 0.0045588 \, J = 4.5588 \, mJ \approx 5.1 \, mJ \]

### Final Answer
The energy that each pulse must deliver to the blemish is approximately **5.1 millijoules**.

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