What is the wavelength of emitted light as an electron in a hydrogen atom goes from N=3 to N=2?
409248
u can find out by\n\n   1/wavelength  =  R(1/(n2)^2 - 1/(n1)^2}\n    and substituting n1=3,n2=2,R(rydberg constant)= 10973731.6\n    you get wavelength in meters then find the corresponding colour of light
1113461
It's 0.00002777778 inches, or approximately 705.55 nanometers.  It's an inch divided into 36000 parts, so it is pretty small.\n\nI believe the wavelength of red light falls in a range of 630 to 760 nanometers or approximately 1/36000 of an inch.
742739
The binding energy of an individual electron in Cs metal is given by the molar work function divided by Avogadro's number:\n\nE_e = 184 kJ/mol / 6.022*10^23 mol^-1\nE_e = 3.055*10^-19 J\n\nThe energy of a photon with wavelength 0.628 nm is given by:\n\nE_p = h*c/lambda\n\nwhere h is Planck's constant\nc is the speed of light in a vacuum\nlambda is the wavelength\n\nso, in this case,\n\nE_p = (6.626*10^-34 (kg*m^2)/s * 2.998*10^8 m/s)/ 0.628*10^-9 m\n\nE_p = 3.163*10^-16 J\n\nE_p exceeds E_Cs by a large margin, so the incoming photon has more than enough energy to eject an electron from the material.  Any energy imparted to an electron that is in excess of the amount needed to eject the electron from the metal will show up as kinetic energy of the electron.\n\nAn electron in a Cs atom beneath the surface of the matrial will require more energy to be ejected than one closer to the surface because the subsurface electron needs additional energy to make its way to the surface.  A electron on a Cs atom at the very uppermost surface will require the least energy to be ejected, and that energy is equal to its binding energy, and it will therefore have the largest "excess" energy, largest kinetic energy, and highest speed.\n\nIn this case, the "excess" energy (Call it E#) is:\n\nE# = E_p - E_e = 3.163*10^-16 J - 3.055*10^-19 J = 3.160 * 10^-16 J\n\nThis energy shows up as kinetic energy:\n\nE# = 0.5 * m * v^2\n\nwhere m is the mass of an electron\nv is the speed of the electron\n\nIn this case:\n\n3.160 * 10^-16 J = 0.5 * 9.109*10^-31 kg * v^2\n\nv = 2.634*10^7 m/s  (I get a slightly different number than the one you were given, which I attribute to rounding differences)\n\nThis speed is 2.634*10^7/2.998*10^8 = 0.088 = 8.8% that of the speed of light in a vacuum.
580324
HAHAHAH ARE YOU TRYING TO TEST US?\n\nLight is a form of electromagnetic radiation: i.e. it has oscillating electric(E) and magnetic(B) fields that travel through  space at speed c with the same frequency and wavelength.\nrotational energy of light if I am not worng...I like that sh.t!!good question\nRotational energy level gaps typically\n\n \n\nDE = 10^-23 J\n\nTherefore n = 1.5 x 10^10 s^-1
750681
A photon with a wavelength of about 650 nm
689936
electron is Negatively charged particle that revolves around the nucleus of an atom.IT shows dual character i.e; it has both wave character and particle character...one electronic charge is 1.609 * 10^-19......the wave length of an electron can be found using the formula lamda= h/mv..where h is the plank's constant 6.626*10^-34...m is mass of an electron 9.1*10^-31...v is its velocity...when supplied energy it gets excited and moves to high enegry orbital...then slowly radiate the energy and then move to their position...these radiations araethe colours seen when we heat substances...the color is due the wavelength of the radiations....the energy of an electron at infinity is 0...when it is near the nuclues...it does some work and energy is spent therefore near the nucleus he charge of an electron is -ve...thats all i can  remember...hope you slect mine...thanks....
46736
Find the number of protons. Then number of electrons = number of protons.
1088371
As far as I know, all of them.  Emission spectra are caused by electrons absorbing energy and moving to a higher energy orbit or state (in the case of visible light it is a higher orbit) then when the electron goes back down to the lower state that it usually inhabits, it gives off a photon of an energy equal to the change in the energy level of the electron.  All elements can have excited electrons, and all electrons give off photons when they return to ground state, so all electrons give off emission spectra.\nIf you know of some element out there that doesn't I would be really interested to know what's going on with it.
459470
they are within the visble range;\nThe series of lines in the hydrogen spectrum associated with transitions to or from the first energy level or ground state. It lies in the ultraviolet (Lyman-alpha at 1215.67 Å; Lyman-beta at 1026 Å) with a series limit is at 912 Å. The He II Lyman lines have almost exactly one-quarter the wavelength of their hydrogen equivalent (He II Lyman-beta, 303.78 Å; He II Lyman limit, 227 Å.)
924166
yes u can. use the equation E = hf where E is the energy and h is the planck constant. once u find f which is the frequecy...insert into equation velocity = wavelength x frequency...assuming u noe the velocity of the hydrogen!! :)\nOH but by the way since ur answer is in kj/mol u need to convert it to kj with the avogadro number which i forget...i'm sure u can find out!
30961
Since the spectra of a given element shows a discrete number of color bands (rather than a continuous specturm), and since the color of light relates to its energy, and since photons are released by the relaxation of electrons from a higher to a lower energy state, it follows that the gaps between energy levels are distinct and finite.  This means that the energy levels are quantized.
960070
Two dimensions should be explanation enough. Assume the de Broglie wavelength of an electron is some distance in nanometers. If we draw the nucleus as a dot and draw the electron as a continuous wave around the nucleus, the wave has to be the length of some multiple of its wavelength. If the wave were not the length of a discrete multiple of its wavelength it would not be continuous because a full circle's rotation would cause the wave to jump. If we constrain the atomic model to behave like this, then the answer is simple. \n\nConsider the electron wave that is four wavelengths long. This wave is a function which describes electron density. The next energy level has to correspond to a wave that is five wavelengths long. In this way, the wave cannot increase gradually, but rather the levels are split into discrete shells, one for each number of wavelengths in the wave.
839956
It emits two lines \nD1 at 589.592 nm\nD2 at 588.995 nm\n\nboth in yellow region of visible spectra
532352
the de Broglie wavelength is the wavelength associated with the matter waves.\n                    λ   =   h/mv\nwhere λ is the wavelength\n      h is the planck's constant = 6.63 X 10^-31 Js\n      m is the mass of the particle\n      v is the speed of the particle.\nIt is valid for all particles in this universe.
863329
N2+3H2=2NH3\n\nFirstly elemental symbols start with capital letters\n\nSecondly you need to know what they mean\nN2 = molecule of nitrogen\nH2 = molecule of hydrogen\n\nThirdly you need to know that NH3 is called ammonia\n\nThen it says 1 molecule of nitorgen + 3 molecules of hydrogen combine to form 2 molecules of ammonia (NH3)
360055
For blue light'\nd Sin th = 2W        W is wave length\nFor a minimum \nd Sin th = (2n+1)W'/2\nSo,  2W=(2n+1)W'/2\nW'=4W/(2n+1)\n  =1440/(2n+1)\nBy putting various values for n , I find that only for the valu n=1, W' is within the vivible range,\nW'=480 nm
78735
Light, form of electromagnetic radiation similar to radiant heat, radio waves, and X rays. Light consists of extremely fast oscillations of an electromagnetic field in a particular range of frequencies that can be detected by the human eye. Different colour sensations are produced by light vibrating at different frequencies, ranging from about 4 × 1014 vibrations per second for red light to about 7.5 × 1014 vibrations per second for violet light. The visible spectrum of light is usually defined by its wavelength, ranging from the smallest visible wavelength for violet, about 40 millionths of a centimetre (16 millionths of an inch), to 75 millionths of a centimetre (about 30 millionths of an inch) for red. Higher frequencies, corresponding to shorter wavelengths, comprise ultraviolet radiation, and still higher frequencies are associated with X-rays. Lower frequencies, which are at longer wavelengths, are called infrared radiation, and still lower frequencies are characteristic of radio waves. Most light comes from electrons that vibrate at these high frequencies when heated to a high temperature. The higher the temperature, the greater the frequency of vibration and the bluer the light produced.\nLight is emitted from a source in straight lines and spreads out over a larger and larger area as it travels; the light per unit area diminishes as the square of the distance. When light strikes an object, it is either absorbed or reflected; light reflected from a rough surface is scattered in all directions. Some frequencies are reflected more strongly than others, and this gives objects their characteristic colour. White surfaces scatter light of all wavelengths equally, and black surfaces absorb nearly all light. Image-forming reflection, on the other hand, requires a highly polished surface such as that of a mirror.\n\nDefining the nature of light has always been a fundamental problem in physics. The English mathematician and physicist Sir Isaac Newton described light as an emission of particles, and the Dutch astronomer, mathematician, and physicist Christiaan Huygens developed the theory that light travels by a wave motion.\n\nIt is now believed that these two theories are essentially complementary, and the development of quantum theory has led to the recognition that in some experiments light acts like a series of particles and in other experiments it acts like a wave. In those situations in which it travels in wave motion, the wave vibrates at right angles to the direction of travel; therefore light can be polarized in two mutually perpendicular planes (see Optics).\n\nIII  VELOCITY \n\nThe speed of light was first measured in a laboratory experiment by the French physicist Armand Hippolyte Louis Fizeau, although earlier astronomical observations had yielded approximately the right velocity. By the 1970s the speed of light had been measured very precisely as 299,792,458 m/sec (186,282.397 mi/sec) in a vacuum, and could be used to measure large distances by the time it took for a pulse of light or radio waves to reach a target and return. This is the principle of radar and sonar. Accurate knowledge of the speed and the wavelength of light also permits accurate measurement of length. In fact, in 1983 the metre was redefined as the length of the path travelled by light in a vacuum during a time interval of 1/299,792,458 of a second. The velocity of light in air varies slightly with wavelength, averaging about 3 per cent less than in vacuum; the speed in water is about 25 per cent less, and in glass, 33 per cent less.\n\nLight has an important effect on many chemicals. Sunlight, for example, is used by plants to carry out photosynthesis, and the exposure of certain silver-containing chemicals to light causes them to turn dark in the presence of other chemicals, as is the case in photography.