Physics, 08.04.2020 04:24 marcos86

Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1.00×1012 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 5.00 μm in diameter, with the pulse lasting for 4.00 ns with an average power of 2.00×1012 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 this cell during this pulse?

B) What is the intensity delivered to the cell? (in W/m^2)

C) What is the maximum value of the electric field in the pulse?

D) What is the maximum value of the magnetic field in the pulse?

Answers: 2

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Answer from: journeyhile5

a) 80 J

b) $1.02\cdot 10^{21}W/m^2$

c) $2.74\cdot 10^7 V/m$

d) 0.091 T

Explanation:

The relationship between energy and power is

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

where

P is the power

E is the energy delivered

t is the time elapsed

In this problem, we have:

$P=2.00\cdot 10^{12} W$ is the average power of the pulse

$t=4.00 ns = 4.00\cdot 10^{-9}s$ is the duration of one pulse of light

Solving for E, we can find the energy given to the cell during this pulse:

$E=Pt=(2.00\cdot 10^{12})(4.00\cdot 10^{-9})=8000 J$

However, this is the energy spread over 100 cells. So, the energy given to 1 cell is:

$E'=\frac{8000}{100}=80 J$

The intensity delivered by the pulse is given by

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

where

P is the power of the pulse

A is the area over which the power is spread

In this problem:

$P=2.00\cdot 10^{12} W$ is the average power of the pulse

The area is the surface area of the cell, which has the shape of a disk, so its area is given by

$A=\pi (\frac{d}{2})^2$

where

$d=5.00 \mu m = 5.00\cdot 10^{-6} m$ is the diameter of the cell

However, the pulse is spread over 100 cells, so the total area to consider is the area of 100 cells:

$A=100\pi (\frac{d}{2})^2$

So the intensity is

$I=\frac{P}{100\pi (\frac{d}{2})^2}$

And substituting we find:

$I=\frac{2.00\cdot 10^{12}}{100\pi (\frac{5.00\cdot 10^{-6}}{2})^2}=1.02\cdot 10^{21}W/m^2$

The relationship between power of an electromagnetic wave and maximum value of the electric field in the pulse is given by

$P=\epsilon_0 E^2 c$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

E is the maximum value of the electric field

$c=3.0 \cdot 10^8 m/s$ is the speed of light

In this problem, we have

$P=2.00\cdot 10^{12}W$ is the average power of the pulse

Therefore, by re-arranging the equation for E, we can find the maximum value of the electric field of the pulse.

It is:

$E=\sqrt{\frac{P}{\epsilon_0 c}}=\sqrt{\frac{2.00\cdot 10^{12}}{(8.85\cdot 10^{-12})(3.0\cdot 10^8)}}=2.74\cdot 10^7 V/m$

In an electromagnetic wave, the relationship between amplitude of the electric field and amplitude of the magnetic field is given by

E=cB

where

E is the amplitude of the electric field

B is the amplitude of the magnetic field

c is the speed of light

Here we have:

$E=2.74\cdot 10^7 V/m$ is the amplitude of the electric field for this pulse

$c=3.0 \cdot 10^8 m/s$ is the speed of light

Solving the equation for B, we find the maximum value of the magnetic field in the pulse:

$B=\frac{E}{c}=\frac{2.74\cdot 10^7}{3.0\cdot 10^8}=0.091 T$

Answer from: Quest

when it gains or loses electrons

Answer from: Quest

Earthplanet yplanet y5=earth *many different equationsplanet x1.5=earthplanet x is two times planet yobject 5,2500 mass5,1,3,2,4object 1 is four times obect 4

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