the photoelectric effect's cut off frequency?
341004
i think you are right, you only need one.\n\nw = h*c/wavelength-eV\n\nboth should give the same w, about 1.9eV, \n\nwavelentgh = h*c/w
411386
photoelectric effect can be demostrated if u can excite an electron in a material so that it has enough energy to leave the matrial. i.e. E > (Evacuum-Ef). now u need to look at the energy of the free electron which can be found simply by putiing in through a magnetic field. the deflection will depend on the velocity and hence its energy. this excitation of the electron is done by electromagnetic radiation or light. this free electron is actually free and is not as that meant in semiconductors where free means that it is STILL in the material but free to move. moreover the "free" electron in a semiconductor, even if excited to a higher energy than the conduction band edge, will come down to the conduction band edge by various non-radiative process and than move. photoelectric effect can also be seen in semiconductors if the excitation is so high such that the electron leaves the material. this method is used to find the position of the fermi level.
910705
Hello, here is what you need:\n\nEnergy = Planck Constant * Frequency\n\nE = h * f\n\nh = 66.26 x 10^-34 Joule second
44523
the speed of light(radiation) is a constant and equals the wavelength times the frequency. As wavelength gets longer the freq goes down and vice versa.
1009361
y dont u do ur homework urself .ok heres abrief intro.\n\n1.see the level of frequency changes when beam of ultra violet light rays enter to one medium from another medium(to glass from air).\n\n2.when light undergoes refraction it frequency changes its speed acc. to the medium such as(glass,air and water)\n\n3.yes there will be,as light enters to one medium from another medium.
283860
Frequency = speed of light/wavelength.\n\nMake sure you keep the units straight when performing the calculations or it may not make sense.\n\nSo for your problem its (3 x 10^17 )/200 = 1.5 x 10^15 Hz
576638
La detectacion de luz ,como photon o como onda, depende del typo o metodo de experimento para explorar la naturaleza de la luz. La luz exhibe la characteristica de honda en experimentos de diffraccion e interferencia y exhibe characteristicas de particula en experimentos de ionizacion de atomos con radiacion de varias energias. La luz se mueve en linea recta a menos que es reflejada o desviada por el medio atraves cual viaja. La luz que proviene de estrellas y galaxias distantes provee informacion soble la velocidad de esas estrellas y galaxias porque la luz como onda es cambiada como es cambiado el sonido de un tren qhe primero se aproxima y despues se aleja. ( vea terminos como "red-shift" and "doppler effect") El sonido aumenta cuando el tren viene y despues mengua cuando el tren se va.
703688
http://submarine.dyndns.org/~tofu702/Science/Physics_Book/PhysicsBookFull.pdf\n\nInduced absorption and stimulated emission in the ground state of a three-level atom\nC Mavroyannis 1992 Quantum Opt. 4 145-150   doi:10.1088/0954-8998/4/3/001\n\n   \n     PDF (237 KB) | References | Articles citing this article \n  \n   \n\nC Mavroyannis\nSci. Inst. for Molecular Sci., Nat. Res. Council of Canada, Ottawa, Ont., Canada\nAbstract. Studies the competition between absorption and stimulated emission induced by two laser fields at low intensities interacting on a two-photon resonance with a three-level atom. The induced spectrum into the ground state consists of a peak whose height and asymmetry depend strongly on the detuning. The spectra arising from the emission of a laser photon by the excited state consists of two peaks describing stimulated emission and induced absorption, respectively, where the former dominates the latter for small values of the Rabi frequencies but for higher ones, they cancel each other out. \n\nPrint publication: Issue 3 (June 1992)\n\n\nPDF (237 KB) | References | Articles citing this article
522376
I believe it DEPENDS on the Frequency of that photon.\nWHITE lite comes into a prism (water droplets) and the\nindividual photos are BENT (refracted) various amounts\ndue to their frequency... thus we get to see the spectrum\n... rainbow... \nSo, yes, a photon going from one medium into another\nwould alter it's path a little at the juncture, \naccording to its frequency...  no "references", \njust my  off-the-cuff reply.  good luck.
482392
It is the range of frequencies from about .7-.4 micrometres.\nWhite light is not a frequency it is the combination of all frequencies in the visible spectrum. The cones in our retina are sensitive to these frequency differences while the rods are sensitive only to the intensity (black & white). The pupil does not respond to light outside of the visible frequencies, therefore infrared and ultraviolet light can easily damage our retinas.
937514
wavelength = speed of light / frequency\nfrequency = speed of light / wavelength\nf = (3*10^8 m/s) / (615*10^-9 m)\nf = 4.878*10^14 Hz\n\nwavelength = planck's constant * speed of light / energy\nenergy = planck's constant * speed of light / wavelength\nE = (6.626*10^-34 J*s) * (3*10^8 m/s) / (615*10^-9 m)\nE = 3.232*10^-19 J
422440
the wavelength is half...\n\nvavelength and frequency are inversely proportional
355213
In y = A sin Bt, B is the frequency in radians/second, or B = 2pi*f, where f is the frequency in cycles/second, or Hertz.\n\nI assume by "c= 3 x 10^10", you're referring to the speed of light, which is 3*10^10 cm/sec.\n\nf = c / wavelength, so f = 3*10^10 cm/sec / (3*10^-10cm) = 3*10^20 cycles/sec or 3*10^20Hz.\n\nSo B = 2*pi*f = 2*pi*3*10^20 = 6*pi*10^20 radians/sec.  That's about 1.885*10^21 radians/sec.\n\nNote that these equations hold for all electromagnetic radiation, including radio waves, microwaves, IR, light, UV, X rays, and gamma rays.
464204
A frequency oscillator.
506358
an energy oscillation
1107375
E=h*f \nc=f*lambda\nf=c/lambda\n\nf(Xray)=(2.9979*10^8)/(5.3*10^-10)=5.656*10^17 [Hz]\nf(visible)=(2.9979*10^8)/(5.3*10^-9)=5.656*10^16 [Hz]\n\nE(Xray)=6.626*10^-34*5.656*10^17=3.747*10^-16 [J]\nE(visible)=6.626*10^-34*5.656*10^16=3.747*10^-17 [J]\n\nSo the visible light is 10 times weaker than the xray.
733136
All elements when heated give off a signature in the light spectrum. For example if you heated a piece of pure iron until it glowed then put a prism in front  of the light, you would see the spectrum of light -- obvious stuff. However because it is the metal iron there would be slots-- frequencies that would be missing. These signatures are all very different from each other depending on which element(s) are giving off light. When something is red shifted the signature is still there but it's place in the spectrum is in the wrong place. It has shifted to the red end of the spectrum-- it is in the longer or red end of it.\n\nThere is two ways to think of this, the other is that when heated elements will give off light in only so many frequencies- the rest of the frequencies are missing.