The unit for measuring the rate at which light energy is radiated from a source is the?
957818
I work with light bulbs and the light output is measured in C. lumens.
747451
c) the frequency of incident radiation\nThis is according to the Einsteins photoelectric equation\nKmax = hf - phi\nwhere phi is the work function, h is the planck's constant and f is the frequency
740581
light, for example is electromagnetic. The absorbance term for light, although dimensionless is defined as -log(T) where T is the % transmittance through the sample.
796397
It's several million, but it is the lens and reflector that do all the work.\n\nLight is measured in lux or illumination is, candle power is the equivalent of Mega  Tonnes when measuring nuclear yield.\n\nYou may find something here\nhttp://hypertextbook.com/facts/2005/DariaZainullina.shtml\n\nSome have 1000Watt bulbs which is only 10 times more than my bathroom light bulb.
889084
E= Energy in Joules\nh= Planck's Constant = 6.626*10^-34 J*s\nv= frequency (Hz or s^-1)\nc= speed of light = 3.0 * 10 ^8 m/s\nx= wavelength of light (given in meters)\n\nTo do this problem you must first convert your wavelength from micrometers to meters. This involves dividing 4.26 micrometers by 10^6, because thats how many micrometers are in a meter.\n\nx= 4.26/(10^6) = 4.26 * 10^-6 meters\n\nAfter this you can plug this value back into the equation to get the amount of energy associated with this photon.\n\nE= ((6.626*10^-34 J*s)(3.0 *10^8 m/s))/ (4.26 * 10^-6 m) \n   = 4.66619718 * 10 ^ -20 Joules
465629
All light has a velocity of "c."  c is ~ 3*10^8m/s.  With this knowledge you can say that frequency is the inverse of time/distance.  Thus, you can use the formula C(meters/sec) / Freq.(cycles/sec) = Wavelength(meter/cycle) to solve the question.\n\nDo your own work from there.
479233
The energy of the incoming photons is given by Planck's relationship:  energy = Planck's constant * frequency.\n\nYou are given the wavelength, not the frequency, so you need to know that for any wave, the wavespeed = frequency * wavelength, so frequency = wavespeed/wavelength.  For photons, wavespeed = c.\n\nh = 6.626068 × 10^-34 34 m^2 kg / s (Planck's constant)\nc = 2.998 × 10^8 m / s (speed of light)\nwavelength = 48.2 x 10^-9 m (given in problem)\n\nPutting this together, the energy of each incoming photon is:\n\nE_photon = h*c/(48.2 x 10^-9 m) = 4.121 x 10^-18 J\n\nThe outgoing electron has kinetic energy = 1/2 * m * v^2.  You are given both m and v:\n\nE_electron = 1/2 * 9.109 x 10^-31 kg * (2.371 x 10^6 m/s)^2 \nE_electron = 2.56 x 10^-18 J\n\nThe ionization energy is the difference in energy between the incoming photon and the kinetic energy of the outgoing electron.  Because the question asks for the energy in joules per mole, we must multiply the energy difference for a single photon/electron pair by Avagadro's number:\n\nE_ionization = (E_photon - E_electron)* N_A\n\nE_ionization = (4.121 x 10^-18 J -2.56 x 10^-18 J)*6.022 x 10^23 mol^-1\n\nE_ionization = 9.4 x 10^5 J/mol = 940 kJ/mol
139259
I'm not sure what you are asking.  Here are two possible answers:\n\nFirst:  E = h*nu\n\nIs the equation for the amount of quantized energy in light where h is Planck's constant, 6.626e-34 Joule seconds, and nu is the frequency in Hertz.\n\nThis determines the "raw" energy in the light.  As you know, going from electricity back to light varies in efficiency.  An electric light bulb requires much more electricity than a flourescent bulb to produce the same amount of light.\n\nSecond: hf = phi + Ek\n\nThis is the equation for the photoelectric effect, where electrons are ejected from a material being struck by light.  h is again Planck's constant and f is the frequency of the incident photon.  Phi is the amount of energy required to remove an electron from its binding and Ek is the kinetic energy observed.
208377
Used for measuring light intesity. It has a photo-voltaic cell(s) that produce electrity when exposed to light. The amount of electricity is measured to find the intensity of light.
210187
Simply put:  it is the luminous intensity expressed in candelas. ( the brightness of light compared to a candle)\n\nA candela is a measurement of the intensity of light, used in the International Standard (SI) measurement system. In the modern age, it is technically defined as the intensity in a given direction of a source that emits monochromatic radiation of a frequency of 540 x 10^12 hertz and which has a radiant intensity in the same direction of 1/683 watts per steradian.\n\nThe candela as a unit of measurement grew out of the desire to adopt a standard method to measure light. In the 19th century, each country had its own measure of light, most extremely difficult to reproduce with accuracy. At the beginning of the 20th century, groups of nations began adopting new standards, such as the international candle, which were based on a single type of lamp and much more consistent. Eventually, in the 1930s, it was decided that a more precise definition was needed.\n\nInitially, the candela was defined as the luminous emission of a Planckian radiator, a type of blackbody, at the temperature of freezing platinum (2045 K). This correlated roughly to the light emitted by a typical candle, making it a suitable measure. In the late 1970s, it was determined that the experimental difficulties in creating a Planckian radiator at such high temperatures made the existing definition of candela less than desirable. The new breakthroughs in radiometry allowed scientists to have a more specific definition, and so the current hertz/watts definition of candela was adopted.\n\nOne reason the watt was not initially tied to the candela is the eye’s differences in seeing various wavelengths of light. Since the human eye is less observant of blue and red light, more wattage is needed to produce the same result in the brain as with yellow or green light. This is why the definition of candela is defined as a very specific hertz, unachievable in actual common lighting. There are those who critique the modern definition of the candela for this reason, arguing that it has removed any common-sense understanding of the candela as an actual emission of light.\n\nThe reason the modern definition of candela includes the somewhat bulky 1/683 watts per steradian is to make it conform to the previous definition, in order to avoid any confusion when comparing the modern unit to the historical candela. In this context, the steradian is the cone of light as it emerges from the source, such that it would light up one meter square of the inside of a sphere with a one meter radius.\n\nSince the precise definition is so unwieldy, it is not uncommon in standard usage to see the candela referred to as “roughly” the amount of light generated by a single candle. This definition is perfectly suitable for common usage, as is the translation of 120 candela to roughly the light emitted by a 100 watt light bulb.
1185872
I can think of two ways to measure the absorption and reflection coefficient of a material.  The first would only work with a polished surface.  First measure the intensity of a laser beam with a photodetector, then shoot the laser beam at the surface and aim the reflection at a photodetector.  Coefficient of reflection = transmitted intensity/initial intensity  (coefficient of absorption =1-reflection).  The other, harder way, which would work with any material, is to measure the heat gain in the target.\n\nFor this you're gonna want as small a piece of the target material you can get so that the laser still completely hits it.  Now attach a thermistor to the back of the target (again the smaller and thinner the target the better, ideal would be a foil just thick enough to be completely opaque).  Now shoot the laser at the target and measure the temperature as a function of time  It should be linear up until it starts radiating nearly as fast as it absorbs energy, so take points until it starts to level off and drop points after that.  Now find the specific heat of the thermistor-target apparatus by dipping it in water with a thermometer in it, using the initial and final temperature of the water, the volume of the water, and the initial temp of the apparatus, determine the heat it absorbed per degree it's temperature changed (this will actually give you specific heat*mass which is fine).  Now convert your temperature vs. time data to heat vs. time using the specific heat you determined (temp*(specific heat*mass)).  The slope of the line of best fit for this data will give you the rate it absorbed energy in watts.  Divide this by the lasers output in watts and you get the absorption coefficient.
470586
E is normally expressed as joules (kilogram-meters per second^2)\n\nSo the mass is in Kg, and the speed of light is in Meters/Sec
872993
E=m c^2\nEnergy=Mass &#92;times (speed of light)^2
963067
Emission lines of different wavelength depending upon what is being studied.
446087
It is the energy emitted per unit time; the power. \n\nTherefore, its dimensional formula is Mass x Length^2 / Time^3.\n\nOr ML^2T^(-3).
538754
hi dude!\ni'll give u only the unworked answers\nwavelength of light i the cornea is 193/1.376 nm\nand the frequency would be 3*10^8*1.376/(193*10^-9) hz\ni hope i'm not too late!!
471190
Electromagnetic radiation in the 1 to 1000 m wavelength range.