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Bob D1
Lv 7
Bob D1 asked in Science & MathematicsPhysics · 8 years ago

E = mc^2 and E = hf : what is the fundamental difference, if any, between ...?

the two forms of energy? Is the energy for mass (E = mc^2), the same or different for photon frequency (E = hf)?

7 Answers

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  • Anonymous
    8 years ago
    Favorite Answer

    1 tells you the energy content as a function of mass whereas the other tells you the energy content as a function of frequency(or wavelength). DeBroglie set them equal to each other and determined that mass has a wavelength.

  • ?
    Lv 7
    8 years ago

    It is best to start from the most general relativistic expression relating energy E to mass m and momentum p:

    E = sqrt( m^2c^4 + c^2 p^2)

    If a mass is stationary with respect to the observer (a massless particle can never be stationary to any observer, it will move at speed c relative to all observers), then for that observer the energy is just the square root of the first term: E = m c^2 .

    A massless particle will have energy E = c p

    Now in th quantum theory of light, it is the combination of the Broglie's p = h / lambda and the standard relation lambda = c / f that together render E = c (h/(c/f) ) = h f.

  • Thomas
    Lv 4
    8 years ago

    Photons have no real mass, yet they have energy. E=mc^2 does not apply in all cases, and isn't actually the exact formula that takes into account relativistic mass. The real E=mc^2 equation is more complex, and is a function of an objects velocity. When that equation is used, it is exactly equal to the E=hf forumula; but keep in mind E=nhf is generally more appropriate, where n is the number of particles.

  • 4 years ago

    they are the two a similar. (it incredibly is the reflexive supplies of equations). E=E. E=one million/2mv^2 describes how a lot speed transformations in a given extraction of however potential a mass has to grant. occasion: a 1kg ball vacationing at 100mps hits the wall. How a lot potential is given off while it hits the wall? preliminary is 100mps, very final is 0. We use a hundred-0 =0 for v. If we mandatory to extract basically area of the balls potential then we'd basically sluggish it down. V subsequently could be a hundred-50 = 50mps (say). the 1st eqn E=mc2 pressumes a similar ingredient, yet utilising c as a widely used speed. the sole distinction between the two is that mc2 is the "all or no longer something" version of one million/2mv^2.

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  • 8 years ago

    I think E = mc^2 and E=hf are related by properties of waves

    (The relation between wave speed, wavelength, and frequency)

    [Below is some algebra that show that the units between E = mc^2 and E = hf agree]

    Wave speed expressed in terms of wavelength and frequency is this

    v = ƛf

    where v: velocity of wave; ƛ: wavelength; and f: frequency

    'c' is the speed of light and substituting this into the wave equation we have

    c = ƛf

    solve for 'f' you get

    f = c / ƛ

    Substitute this into E = hf you get

    E = h * (c / ƛ)

    Now replace this with the 'E' in E = mc^2 to get

    h * (c / ƛ) = mc^2

    h / ƛ = mc

    c = h / ƛ *m

    where 'h' is measured in joule seconds (J*s)

    The units of "h / ƛ *m" are

    J*s / m * kg

    ([kg*m^2] / s) * (1/kg * m)

    = m / s

    So the units of "h / ƛ *m" are in meters per second which agrees with the units on the left side for 'c' the speed of light which should also be measured in meters per second.

    Source(s): _____________________________ I think the answers above are more informative than mine.
  • 8 years ago

    They are equal.

    mc^2 = E = hf

    In this case m refers to the relativistic mass, not the rest mass. Light has no rest mass, but it does have relativistic mass, which can be solved by the above equation.

    E^2 = (m0c^2)^2 + (pc)^2 is also true. m0 is the rest mass.

    (mc^2)^2 = (m0c^2)^2 + (pc)^2

    For light

    (mc^2)^2 = (pc)^2

    mc^2 = |pc|

    |p| = mc

    Since |v| = c for light, the Newtonian formula for momentum, p = mv, works for light too by coincidence, when m is relativistic mass.

  • Anonymous
    8 years ago

    No, human on planet earth are not counting speed of light and speed of sound using true physic, because if you count with true physic you'll get the speed are equal at light and sound, but when you count with biological and medical disabilities you'll get light is faster then sound.

    another explanation is devices to aid collective signals of light is easy, but devices to aid collective signals of sound is hard.

    so what is devices to aid scientist to collect light and why it is easy, also what is devices to aid scientist to collect sound and why it is hard?.

    devices to aid scientist to collect light need only to be sensitive to heat and burned, but devices to aid scientist to collect sound need to be thinner and thinner and thinner and vibrates..

    Source(s):

    The Aliens (Extra terrestrial) that gave me ideas, want me to give all to the present Government of Planet Earth.

    5 seconds after a Flash 1 mile away

    10 seconds after a Flash 2 miles away

    15 seconds after a Flash 3 miles away

    20 seconds after a Flash 4 miles away

    25 seconds after a Flash 5 miles away

    30 seconds after a Flash 6 miles away

    35 seconds after a Flash 7 miles away

    40 seconds after a Flash 8 miles away

    Source(s):

    http://www.lightningsafety.noaa.gov/distance.htm

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