Question

High-Energy Cancer Treatment. Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of ${\mathbf{10}}^{\mathbf{12}}\mathbf{}\mathbf{W}$) pulses of light that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 5.0 mm in diameter, with the pulse lasting for 4.0 ns with an average power of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{12}}\mathbf{}\mathbf{W}$. We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse. (a) How much energy is given to the cell during this pulse? (b) What is the intensity (in ) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

Short answer

a. 8 kJ of energy is given to the cell during pulse.

b. The amount of intensity delivered to cell is ${10}^{21}\mathrm{W}/{\mathrm{m}}^2$.

c. The maximum values of the electric and magnetic fields in the pulse are $8.7\times {10}^{11}\mathrm{V}/\mathrm{m}$and $2.9\times {10}^3\mathrm{T}$respectively.

Step-by-step solution

Define the intensity ( I  ) and define the formulas.

The power transported per unit area is known as the intensity ( I ) .

The formula used to calculate the intensity ( I ) is:

$I=\frac{P}{A}$

Where, $A$is area measured in the direction perpendicular to the energy and$P$ is the power in watts.

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

$$\begin{gathered} E_{max}=\sqrt{\frac{2I}{{\epsilon }_{0}c}} \\ {B}_{max}=\frac{E_{max}}{c} \end{gathered}$$

Where,${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N·m^2$ and c is the speed of light that is equal to $3.0\times {10}^{8}m/s$.

The energy of flows through the area is related to time as $\left(F\right)=Pt$.

Determine the energy.

Given that,

$$\begin{gathered} P=2\times {10}^{12}\mathrm{W} \\ t=4\times {10}^{-9}\mathrm{s} \end{gathered}$$

The energy of flows through the area is related to time is

$\left(F\right)=Pt$

Substitute values in above equation

$$\begin{gathered} F=\left(2\times {10}^{12}\right)\left(4\times {10}^{-9}\right) \\ =8\mathrm{kJ} \end{gathered}$$

Hence,$8\mathrm{kJ}$ of energy is given to the cell during pulse.

Determine the intensity.

Given that,

$$\begin{gathered} P=2\times {10}^{12}W \\ d=5\mu m \end{gathered}$$

The formula used to calculate the intensity$\left(I\right)$ is:

$I=\frac{P}{A}$

Substitute the values

$$\begin{gathered} I=\frac{2\times {10}^{12}}{\pi \left({\displaystyle \frac{5\times {10}^{-6}}{2}}\right)} \\ =\frac{2\times {10}^{12}}{19.63\times {10}^{-12}} \\ ={10}^{21}\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the amount of intensity delivered to cell is ${10}^{21}\mathrm{W}/{\mathrm{m}}^2$.

Determine the maximum values of electric and magnetic fields.

The maximum value of electric field is:

$E_{max}=\sqrt{\frac{2I}{{\epsilon }_{0}c}}$

Substitute the values

$$\begin{gathered} E_{max}=\sqrt{\frac{2\times \left({10}^{21}\right)}{\left(8.85\times {10}^{-12}\right)\left(3\times {10}^8\right)}} \\ =8.7\times {10}^{11}\mathrm{V}/\mathrm{m} \end{gathered}$$

The amplitude of magnetic field is:

$B_{max}=\frac{E_{max}}{c}$

Substitute the values

$$\begin{gathered} B_{max}=\frac{8.7\times {10}^{11}}{3\times {10}^8} \\ =2.9\times {10}^3\mathrm{T} \end{gathered}$$

Hence, the maximum values of the electric and magnetic fields in the pulse are$8.7\times {10}^{11}\mathrm{V}/\mathrm{m}$ and$2.9\times {10}^3\mathrm{T}$ respectively.