Energy per photon = ( h ) times (frequency)
( h = Planck's Konstant = 6.63 x 10-34 joule-second )
Frequency = speed/wavelength
Energy = ( h ) x (speed/wavelength) = (6.63 x 10-34) x (3 x 108/4.6 x 10-7) = 4.32 x 10-19 joule
4.44 10-19 j
400 nm= 400 × 10-9 m
C= λv
3.00 x 108 m/s= 400 × 10-9 m (v)
v= 7.50 × 1014 Hz
E=hv
E= 6.626 × 10-34 J (sec) × (7.50 × 1014 1/sec)
E= 4.97 × 10-19 J
The math above follows you through the steps of finding the energy (in Joules) from a specified wavelength. In this situation, it's best to convert to meters to accommodate the speed of light measured in meters per second further down. (Of course, it's at the physicists discretion to change the units of the speed of light instead). Then you go through a two-step process: Finding the frequency (v) and using the frequency to find the energy (E). To find the frequency, use the formula C= λv where C = the speed of light and λ (lambda) = the wavelength. We know two of the variables (speed of light and wavelength), so we're ready to find the third variable, frequency, which simple arithmetic should provide for you.
Then, knowing the frequency, we can enter the value into the second equation, E=hv, where E = energy (what we're finding) and h = Plank's constant (it may or may not be necessary to know what this represents depending on the level of your course). Again, we plug in the frequency and the constant to find E, energy.
Also, a photon with a wavelength of 400 nm will have an energy of 3.1 eV. (This might actually be called the work function associated with that wavelength.) And its more purple or violet than blue, by the way. To convert wavelength to energy (work function) for a given wavelength, divide 1240 by the wavelength in nm to discover the answer.
See related link below for a corroboration from Wolfram|Alfa.
The answer is 4.4 x 10-19 J, or 2.75 eV.
Here's how I got that number:
The energy of a photon is given by the following equation, E = hf, where E is energy, h is Planck's constant, and f is the frequency of the photon.
Additionally, the speed of a photon of light is given by the following equation, c = fλ, where c is the speed of the photon, which is constant in a vacuum, and λ is the wavelength of the photon.
Therefore, f = c/λ.
Substituting this expression for f into the equation E = hf, gives us the equation E = hc/λ.
Now, as I said before, c is a constant, but so is h. Their values are 3.0 X 108 m/s and 6.63 x 10-34 Js, respectively. Thus, plugging in the value in question for λ, 452 nm, or 452 x 10-9 m, we get:
6.63 x 10-34 Js * 3.0 x 108 m/s / 452 x 10-9 m, which equals
4.4 x 10-19 J, or 2.75 eV.
450 nm= 450 × 10-9 m
C= λv
3.00 × 108 m/s= 450 × 10-9 m (v)
v= 6.67 × 1014 Hz
E=hv
E= 6.626 × 10-34 J (sec) × (6.67 × 1014 1/sec)
E= 4.42 × 10-19 J
Eh, this is really simple, open your book. E=hf=hc/lambda.
Use the formula: energy = hc / lambda, where "h" is Planck's constant, "c" is the speed of light, and "lambda" is the wavelength. Don't forget to convert the wavelength to meters first.
(1 eV / 1.6 x 10^(-19))(3.6 x 10^(-19) = 2.25 eV
3.6 x 10-19 J
Ultraviolet radiation has a wavelength between 10 and 400 nanometers. The shorter the wavelength, the more energy each photon contains. To find the frequency, divide the speed of light (299,792,458 metets per second) by the wavelength.
Photon energy can be increased by following two methods. 1). by increase in frequency of one photon as (E = hf); where f denotes the frequency of corresponding region. In this case, the electromagnetic region will change to higher frequency region or shorter wavelength region. The photon energy may increase, but not the intensity. 2). secondly increase in the number of photons (n) as E= nhf. If the number of photons of a particular frequency increase, photon energy also increases. In this case, intensity of light of definite frequency (either blue, red etc.) increase simultaneously.
Purple is a mixture of the colors red and blue.
There are different types of energy. In this case, the energy apart from radiant heat is giving off energy in the visible range, ~400-800 nm wavelength. That's why you can see it.
Wavelength of violet is the shortest and that for red is the longest in the visible region. So for ultra violet the wavelength is to be less still and that for infra red it has to be larger than red So wavelength increases as we move through UV, visible and IR.
Energy per photon is proportional to frequency. That tells us that it's alsoinversely proportional to wavelength.So if Photon-A has wavelength of 400-nm, then wavelength of Photon-Bwith twice as much energy is 200-nm .
Ultraviolet radiation has a wavelength between 10 and 400 nanometers. The shorter the wavelength, the more energy each photon contains. To find the frequency, divide the speed of light (299,792,458 metets per second) by the wavelength.
Photon energy can be increased by following two methods. 1). by increase in frequency of one photon as (E = hf); where f denotes the frequency of corresponding region. In this case, the electromagnetic region will change to higher frequency region or shorter wavelength region. The photon energy may increase, but not the intensity. 2). secondly increase in the number of photons (n) as E= nhf. If the number of photons of a particular frequency increase, photon energy also increases. In this case, intensity of light of definite frequency (either blue, red etc.) increase simultaneously.
There are spaces in the atomic spectrum of hydrogen because there are discrete energy levels that the electron in the hydrogen atom can be located in. Generally speaking the further away from the nucleus, the higher the potential energy of the electron. When hydrogen gas is excited, the electron can jump up to higher energy levels. When that electron falls back down to a lower energy level, a photon is emitted with an energy equal to the energy difference between the atomic orbital it jumped from and the one it jumped to. Since excited electrons can make a number of different jumps (ex. 4->3, 4->2, 5->3, 5->2, etc) there are a series of photons given off with discrete energies. Each one of these photons has a distinct wavelength (given by the equation E=hf, where E is the energy of the photon, h is planck's constant and f is the frequency of the photon). Each line you see on the spectrum is a photon produced from a different energy jump, with a different wavelength. We are only able to see the photons that emit a wavelength in the visible spectrum (roughly 400-700 nm).
Short wave radiation. You might separate it to X-ray or Ultra-Violet but basically, any wavelength shorter than 400 nm had degree of harm. The shorter wavelength the more energy/photon and higher risk of causing DNA damage and resulting to cancer.
(300,000,000 meters per second) / (750,000 waves per second) = 400 meters per wave
Purple is a mixture of the colors red and blue.
It affects the color of the light. If you change the wavelength of visible light, you change the light's color. For instance. Red is a wavelength of approximately 650 nanometers. Orange, about 590. Yellow, 570. Green- 510, Blue, 475 nanometers. Indigo is 445 and Violet is about 400. You can see more here: http://eosweb.larc.nasa.gov/EDDOCS/Wavelengths_for_Colors.html#red
400-490
Radiation is photon or light. We give name for radiation by its' range of wavelength. Those with wavelength < 0.001 nm is called gamma ray, those below gamma to just about 0.1 nm is x-ray and those below x-ray are UV and visible light. Our visible light is light with wavelength 400-700 nm below that are infrered, microwave and radiowave.
There are different types of energy. In this case, the energy apart from radiant heat is giving off energy in the visible range, ~400-800 nm wavelength. That's why you can see it.
blueee!!