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1 year ago

How are electromagnetic waves used in a bar-code scanner?

1 answer:

meriva1 year ago

5 0

The electromagnetic waves used in a bar-code scanner by an image is produced when a silicon chip releases electrons by the photoelectric effect and Visible light is reflected off an object and then captured by a charge-coupled device.

<h3>What is electromagnetic wave?</h3>

The electromagnetic wave is the white light or Sunlight coming directly from the Sun. This light ray is the form of wave. This light when falls on any object segregates into the range of all types of EM radiation.

Radiation is packet of energy that travels and spreads out. It has radio waves, infrared waves and visible light waves, ultraviolet rays and microwaves.

Bar code scanner uses the photoelectric effect and visible light reflection phenomenon.

Thus, the correct options are A and C.

Learn more about electromagnetic wave.

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lcompute the increase in length of 500 m of copper wire when it's temperature changes from 12 degrees celsius to 32 degrees Cels

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Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?

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As we know that the formula of kinetic energy will be

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2 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

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Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

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2 years ago

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belka [17]

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The Weight is,

$F=mg=1948*98=19090.4N$

Through the hook's Law we calculate X.

$F_s=Kx$ , where x is the lenght of compression and K the Spring constant.

We don't have a K-Spring, but we can assume a random value (or simply let the equation in function of K)

$X = \frac{F_s}{x} \\X = \frac{1909.4}{k}$

I assume a value of $K=4*10^4N/m$

$X= \frac{1909.4}{4*10^4} = 0.48m$

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2 years ago