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The photon energy is directly proportional to its frequency: Energy = Planck's constant frequency.

A high energy light will have a shorter wavelength than a low energy light. If the wavelength goes down, then the frequency goes up. When calculating energy in the equation, E=hv, frequency (v) is the variable, not the wavelength. So in the equation, if you wanted a more energy (E), you would have the frequency be large. For the frequency to be big, then the wavelength has to be low.

575nm = 575 x 10 -9 mC= λv3.00 x 10 8 m/s = 575 x 10 -9 m (v)v= 5.22 x 10 14 HzE= hvE= 6.626 x 10 -34 J (sec) x (5.22 x 10 14 1/sec)E= 3.46 x 10 -19 J

The maximum energy of a photon is often expressed as Planck's Lawor constant. It is also referred to as "h" and is used in quantummechanics as well.

2.8x10 -19 J E = hν, and ν = c/λ. λ is wavelength, c is the speed of light (3.0 x 10 8 meters/second), E is energy, and h is 6.626 x 10 -34 . 3.0 x 10 8 /715 = 419580.42. 419580.42 x 6.626 x 10 -34 = 2.78 x 10 -28 joules .

Photopigments release energy when they come into contact with light. This causes a second messenger system in the cell to send the sensation of a certain color of light. the photopigment chlorophyll, when absorbing a photon, gets an electron knocked out of orbit by it. This electron follows an electron transport system which is the first step in the photosynthetic creation of energy in the form of ATP I'm still trying to figure out what happens to the photopigment melanin when it absorbs a photon, so far i can only tell that heat gets dissipated, though there is evidence that different types of melanin will induce apoptosis (early stage cell death) in surrounding cells.

575nm = 575 x 10 -9 m. C= λv. 3.00 x 10 8 m/s = 575 x 10 -9 m (v). v= 5.22 x 10 14 Hz. E= hv. E= 6.626 x 10 -34 J (sec) x (5.22 x 10 14 1/sec). E= 3.46 x 10 -19 J

Energy = hc/λ h = Planck's constant (6.626 × 10 -34 m 2 kg/s) c = the speed of light (3.00 × 10 8 m/s) λ = the wavelength Example question: What is the energy of a photon with a wavelength of 700 nm? Convert the wavelength to meters (1 nm = 10 -9 m) 700 nm = 700 × 10 -9 m Energy = (6.626 × 10 -34 )(3.00 × 10 8 ) ÷ (700 × 10 -9 ) = 2.84 × 10 -19 J

The visible light spectrum ranges from red light (620 nm) to violet light (480nm). Because wavelength is inversely proportional to energy, violet light posseses the most energy while red light possesses the least (as measured in photons).

An electron emits a photon when it lowers its energy, going for a lower energy orbital. The energy of the photon will be the difference of energy between the orbital where the electron was, and the orbital where the electron went.

Those with the highest frequency/shortest wavelength. Roughly speaking, highest to lowest: Gamma or X-Rays, UV Light, Blue light, Red Light, Infra-Red, Microwave, Radio waves.

No. For two reasons: 1- By amount of radiation, you mean Intensity, which is a variable for number of photons. So you can't increase intensity with just one photon; 2- The energy of a photon only depends on its frequency or wavelength.

The amount of energy in a photon of light is proportional to the frequency of the corresponding light wave. ... frequency of the electromagnetic radiation of which the photon is a particle.

Photon Energy E=hf = hc/w thus wavelength w= hc/E or the wavelength is hc divided by the energy of the photon or w= .2 e-24 Joule meter/Photon Energy.

All particles which represents a quantum of light and otherelectromagnetic radiation is called photon. The photons with thehighest energies are gamma or X-rays, UV light, Blue light, andradio waves.

E = photon energy h = Planck's constant c = speed of light F = frequency L = wavelength E = h F which is the same as saying E = h c / L There's your photon energy in terms of wavelength.

Photons with the lowest energy have the lowest wavelength. Thesephotons with lowest energy also have the lowest frequency comparedto higher energy ones.

Within the range of visible light, it's the light with the shortest wavelength (highest frequency). That's the violet end of the spectrum. Past that, ultraviolet light, which is invisible, has higher energy.

The energy of any given photon can be calculate through the equation E= hf, where: E = the energy h = Planck's constant (6.626 x 10^-34 m^2 Kg/s) f = the frequency of the photon

Photons at the red end of the visible spectrum have the longest wavelength and lowest frequency, and carry the least energy. Photons at the violet end of the visible spectrum have the shortest wavelength and highest frequency, and carry the most energy.

It's the source of life and understanding for who we are today, so that we will be able to make sense of who we are within the universe.

You need to know the photon's frequency or wavelength. If you know the wavelength, divide the speed of light by the photon's wavelength to find the frequency. Once you have the photon's frequency, multiply that by Planck's Konstant. The product is the photon's energy.

The rainbow color with the highest energy photons is violet. Therainbow color with the lowest energy photons is red.

what is the energy of a photon that has a frequency of 5.0x10 14 Hz?

you see how close the wavelengths are-- the closer, the more powerful it is. Here are the wavelengths- from least powerful to most powerful: Radiowaves, Microwaves, Infared Waves, Visable Light, Ultraviolet, X-rays, and Gamma rays.

The energy of a photon is directly proportional to its frequency.\n The energy of a photon is directly proportional to its frequency.\n The energy of a photon is directly proportional to its frequency.\n The energy of a photon is directly proportional to its frequency.\n

The energy of a photon is E=hf where h is Planck's Constant 2/3E-33 joule seconds and f is the frequency of the photon in vibrations per seconds.

No. Basically the energy is proportional to the photon's frequency, or inversely proportional to its wavelength.

An X-ray proton. This is so because the x-ray has much higher frequency and shorter waves.

Microwaves have a wavelength of 12.2 cm (abbreviated with a lambda)and a frequency (f) of 2.45 GHz. Einstein discovered that themagnitude of an energy quantum (a microwave photon in this case) isproportional to the wave frequency by the following equation. E = hf Where E is the total energy, h is Planck's constant (h=6.626 10^-34 Js), and f is the frequency. Plugging in the numbers we get E = (6.62610^-34 Js) (2,450,000,000 Hz) = 1.6233710^-24 J

The Energy Of photon is given by E = h ν where h = planks constant ν = frequency of ray as sun rays are combination of various frequency waves, the energy of each photon is dependent upon the frequency or wavelength of rays.

A photon with energy 3.0 x 10-19 J A photon with wavelength 525 nm A photon with frequency 7.6 x 1014 Hz A photon with frequency 2 x 1015 Hz

The name photon derives from the Greek word for light, φως (transliterated phôs), and was coined in 1926 by the physical chemist Gilbert Lewis.

ELF. Radio signals in this band have lower frequencies than the lowest audible sound waves.

Energy content would be measured in Joule. Mole is a measure of "quantity of substance", basically, counting or calculating the number of photons. It is not related to energy content.

i got 3.310^-19... im not positive. so if you need anthing tocheck your work on... that is my solution

The energy of a photon is directly proportional to its frequency. The frequency (and therefore also the energy) are inversely proportional to the wavelength (for any wave, frequency x wavelength = speed of the wave).

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 10 8 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 10 8 m/s / 452 x 10 -9 m, which equals 4.4 x 10 -19 J, or 2.75 eV.

A photon can have just about any energy - from near zero, to extremely high energies. The energy, by the way, is proportional to the frequency.

I assume you want energy per kilojoules a mole since the wavelength is in nanometers. Energy =1.196 X 10^5 nmkj/mol/wavelength E = 1.196 X 10^5 nmkj/mol/715 nm = 167 kilojoules/mol of photons If you just wanted Joules, then this way is recommended. ( change 715 nm to 7.15 X 10^-7 meters ) Energy ( in Joules ) = Planck's constant speed of light/wavelength in meters E = ( 6.626 X 10^-34 Js)(2.998 X 10^8 m/s)/( 7.15 X 10^-7 meters) = 2.78 X 10^-19 Joules

To calculate the energy of a single photon of an electromagnetic wave, you first need to know the frequency of the wave. If you know the value of 'f', you can calculate the value of 'E' by using the formula E = h x f, where 'h' is the Planck's constant. It has a value of 6.63 x 10 -34 m 2 kg/s.

We can use the Plank's Law: E = hf = hc / λ Where h is Plank's constant (7 x 10 -34 J/s)[approx.] c is the speed of light in a vacuum (3 x 10 8 m/s) λ is the wavelength, in meters, of the photon you're measuring E = [ (7 x 10 -34 )(3 x 10 8 ) ] / 3 x 10 -9 m E = [ 21 x 10 -26 ] / 3 x 10 -9 m E = 7 x 10 -19 J

i have no idea but i waz looking that up cuz, i think we might have the same homework question, if u get the answer plz tell me!

Not. Photons don't 'bind' to anything. As soon as they're created, they take off, at the speed of light.

It depends on the wavelength of the photon. Energy of each photon is hc/λ, where h = Planck's constant = 6.626x10 34 Js, c = speed of light = 3x10 8 m/s, and λ = wavelength of the photon

The higher the frequency of the EM radiation, the greater the energy of an individual photon. X-rays have far higher frequencies than do radio waves.

It's proportional to the frequency of the photon ... 6.63 x 10 -34 joule per Hz .

No. But if the photon has a very large wavelength (and a very small frequency) its energy can be very close to zero.

Both of them can have a low or a high energy. . The question is essentially meaningless for the above reason, but let's try to answer it anyway.. Neutrons have an inherent energy equivalent to their mass (which is, for a subatomic particle, pretty substantial).. The energy of a photon can be pretty much any desired value. Low-energy photons have low frequencies; high-energy photons have high frequencies.. The energy equivalent of a neutron at rest is roughly equal to that of a photon with a frequency of 2.2 x 10 23 Hz. This is an amazingly high frequency (way up in the gamma region of the spectrum, which starts at around 10 19 Hz), so it's rather likely that for any given photon/neutron pair, the neutron has a higher overall energy.. Gamma-ray bursts can contain photons with energies far higher than a neutron at rest, though. The neutron energy is around a GeV; gamma-ray bursts can have photons with energies in excess of 10 TeV, 10,000 times higher..

Energy = hf = hc/wavelength Energy = 1.986×10 e −25 / 518 x 10 e -9 = 3.834 x 10 e - 37 J


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