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What is the actual diameter of the red blood cell?

  1. Dec 6, 2006 #1
    need some help please :)

    Had major paper due today, didn't have much time for my physics hw, and now I am stumped on some. Anyone got ideas, I have to turn it in tommorow! Thanks for any help you can give!

    1. A microscope has an objective lens with a focal length of 16.22mm and an eyepiece with a focal length of 9.50mm. With the length of the barrel set at 29.0cm, the diamter of a red blood cell's image subtends an angle of 1.43mrad with the eye. If the final image distance is 29.0 cm from the eyepiece, what is the actual diamter of the red blood cell?

    2. Calculate the energy, in electron volts, of a photon whose frequency is
    a) 620 THz
    b) 3.1 GHz
    c) 46.0 MHz

    d) Determine the cooresponding wavelengths for these photons and state the classification of each on the electromagnetic spectrum



    3. The threshold of a dark-adapted (scotopic) vision is 4 * 10^-11 W/m^2 at a central wavelength of 500nm. If light with this intensity and wavelength enters the eye when the pupil is open to its maxiumum diamter of 8.5mm, how many photons per second enter the eye?



    4. Lithium, beryllium, and mercury have work functions of 2.3 eV, 3.9 eV, and 4.5 eV, respectively. If 400nm light is incident on each of these metals, determine

    a) which metals exhibit the photoelectric effect and

    b) the maxium kinetic energy for the photoelectrons in each case



    5. Xrays with an energy of 300 keV undergo compton scattering from a target. If the scattered rays are deflected at 37.0 degrees relative to the direction of the incident rays, find

    a) the compton shift at this angle

    b) the energy of the scattered xray

    c) the kinetic energy of the recoiling electron
     
  2. jcsd
  3. Dec 6, 2006 #2

    andrevdh

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    1. If we assume that the eye is up against the eyepiece all information needed is the angle that the bloodcell subtends at the eye and the distance that it is from the eye (eyepiece): 29 cm. For such a small angle we can then use the radian measure

    [tex]\theta _r = \frac{D}{29}[/tex]

    where D is the diameter of the cell in cm. That is

    [tex]0.00143 = \frac{D}{29}[/tex]

    http://www.saburchill.com/physics/chapters3/0014.html
     
    Last edited: Dec 6, 2006
  4. Dec 6, 2006 #3

    andrevdh

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    2. The energy of a photon, [tex]E[/tex], is related to its frequency, [tex]f[/tex], by Planck's constant, [tex]h[/tex], according to

    [tex]E = hf[/tex]

    http://www.allrefer.com/planck-s-constant

    The electron-volt is also a measure of energy

    http://www.allrefer.com/electron-volt

    the "unit of energy" just differs from our "normal" energy unit - joule.

    The wave equation can be used for the wavelengths

    [tex]c = \lambda f[/tex]

    using the known speed and frequencies. To identify the type of waves you are dealing with see

    http://www.allrefer.com/pictures/s1/e0077200-electromagnetic-spectrum
     
    Last edited: Dec 6, 2006
  5. Dec 6, 2006 #4

    andrevdh

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    3. The threshold intensity tells us how many photons is needed per second per square meter to be just visible. If one divides this energy value (joules per second per square meter) by the energy per photon you will get the amount of necessary photons per square meter per second to just cause a visible stimuli. The eye will intercept only a fraction of these amount of photons since its pupil has a much smaller area.
     
  6. Dec 6, 2006 #5

    andrevdh

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    4. In the photoelectric effect we find that the energy of the photons is used to liberate photoelectrons from the material

    http://reference.allrefer.com/encyclopedia/P/photelef.html

    A photon needs a minimum amount of energy to just liberate a photoelectron from the material (tear it out of the surface with zero kinetic energy left for the photoelectron). This minimum amount of energy is called the work function of the material. The left over energy (of the photon) will appear as the kinetic energy of the liberated photoelectron.
     
    Last edited: Dec 6, 2006
  7. Dec 6, 2006 #6

    andrevdh

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    http://www.britannica.com/eb/article-9361304/Compton-effect

    http://physics.about.com/od/quantumphysics/a/comptoneffect.htm

    The compton shift of the X-ray photon is given by

    [tex]\Delta \lambda = \lambda _C (1 - \cos(\theta))[/tex]

    where [tex]\lambda _C = 2.426 \times 10^{-12}\ m[/tex]

    this is the amount by which the photon's wavelength increases (energy lost to the interaction with the electron implies increase in wavelength [longer wavelength implies lower frequency]).

    Since we know that the energy of a photon is given by

    [tex]E = hf = \frac{h}{c} \lambda[/tex]

    it follows that its energy change will be given by

    [tex]\Delta E = \frac{h}{c} \Delta \lambda[/tex]

    this will also be the amount of energy lost to the electron (energy conservation).
     
    Last edited: Dec 6, 2006
  8. Dec 6, 2006 #7
    thanks for the help :)
     
  9. Dec 6, 2006 #8

    andrevdh

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    It's a pleasure.
     
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