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We have an equation for that:
lambda = c / (f * sqrt(epsilon))
where lambda = wavelength in [m]
c = the speed of light in vacuum = 3E+8 [m/s]
epsilon = the dielectric constant of the medium in question = 1 for air or vacuum
Hence, the photon frequency in air or vacuum = c / lambda = 3x108 / 0.21 [s-1] = 1.43 [GHz].
The photon energy, E = h * f, where h = Planck's constant = 6.63x10-34 [Js].
E = 6.63x10-34 [Js] * 1.43 [GHz] = 9.5x10-34 [J]
What else can I help you with?
Which color of light does the smallest energy drop of an electron produce?
The smallest energy drop of an electron produces red light. When an electron transitions to its lowest energy level, it emits a photon with the least energy, corresponding to the red wavelength of light.
A typical wavelength of infrared radiation emitted by your body is 25 micrometers. What is the energy per photon of such radiation?
* E = hf = hc/wavelength = (6.63 x 10-34 J*s)(3.00 x 108 m/s)/(25 x 10-6 m) = 7.9 x 10-21 J per photon. This is the energy of a photon at that wavelength. == The person who asked the question answered it. Why ask a question to which you already know the answer? And the body under "normal" conditions radiates infrared (IR) most strongly at about 10 micrometers.
How much energy does a photon of frequency 6 x 1012hz have?
Simply use the formula E = h * frequency h - Planck's constant and its value is 6.626 x 10-34 J s As we plug 6 x 1012 Hz for frequency we get E in joule So E = 3.9756 x 10-21 J
What is the energy of an electromagnetic wave with the frequency of 8E12 Hz?
1.99 10-24 j
When do atoms of cool hydrogen emit 21 cm radiation?
Atoms of cool hydrogen emit 21 cm radiation when their electrons transition from a higher energy state to a lower energy state, specifically when the electron's spin flips from being parallel to anti-parallel to the proton's spin. This transition occurs at a wavelength of 21 cm, which corresponds to a frequency of about 1.42 GHz. The radiation is typically emitted by neutral hydrogen gas in space, particularly in regions of low density and temperature, such as the interstellar medium. This emission is crucial for astronomers to map hydrogen distribution in the universe.
Which color of light does the smallest energy drop of an electron produce?
The smallest energy drop of an electron produces red light. When an electron transitions to its lowest energy level, it emits a photon with the least energy, corresponding to the red wavelength of light.
A typical wavelength of infrared radiation emitted by your body is 25 micrometers. What is the energy per photon of such radiation?
* E = hf = hc/wavelength = (6.63 x 10-34 J*s)(3.00 x 108 m/s)/(25 x 10-6 m) = 7.9 x 10-21 J per photon. This is the energy of a photon at that wavelength. == The person who asked the question answered it. Why ask a question to which you already know the answer? And the body under "normal" conditions radiates infrared (IR) most strongly at about 10 micrometers.
Find the energy of a photon whose frequency is 5x10 12 Hz?
The energy of a photon can be calculated using the formula E = h * f, where h is Planck's constant (6.626 x 10^-34 J*s) and f is the frequency of the photon. Thus, for a frequency of 5 x 10^12 Hz, the energy of the photon would be 3.31 x 10^-21 Joules.
How much energy does a photon of frequency 6 multiply 1012 Hz have?
The energy of a photon is given by E = hf, where h is the Planck's constant (6.626 x 10^-34 J·s) and f is the frequency of the photon. Plugging in the values, the energy of a photon with a frequency of 6 x 10^12 Hz is approximately 3.98 x 10^-21 Joules.
How much energy does a photon of frequency 6x10 12 Hz have?
The energy of a photon is given by the formula E = hf, where h is Planck's constant (6.626 x 10^-34 J s) and f is the frequency of the photon. So, for a photon with a frequency of 6 x 10^12 Hz, the energy would be approximately 3.98 x 10^-21 Joules.
How much energy does a photon of frequency 6 and times 1012 Hz have?
The energy of a photon can be calculated using the equation E = hf, where E is the energy, h is Planck's constant (6.626 x 10^-34 J s), and f is the frequency of the photon. Plugging in the values, the energy of a photon with a frequency of 6 x 10^12 Hz would be approximately 3.98 x 10^-21 Joules.
How much energy in Joules is in one photon of light with a wavelength of 550 nm?
3.04 10-19 j
What is the wavelength and frequency of photons with and energy of 1.410-21?
Let us use the expression E = h v v is the frequency in Hz So v = E / h E = 1.4 x 10-21 J h = planck's constant = 6.676 x 10-34 Js Plug and solve for v So v = 2112.9 G Hz To have wavelength lambda we have to use the expression lambda = c/v c = the velocity of light in vacuum ie 3 x 108 m/s Plug and solving we get lambda = 1.42 x 10-4 m
How much energy does a photon of frequency 6 x 1012 Hz have?
The energy of a photon is given by the equation E = hf, where h is Planck's constant (6.626 x 10^-34 J*s) and f is the frequency. Plugging in the values, the energy of a photon with a frequency of 6 x 10^12 Hz would be approximately 3.98 x 10^-21 Joules.
When was Social Wavelength created?
Social Wavelength was created on 2009-04-21.
What is the wavelength of the hydrogen?
21 centimeter line
What is the energy of a photon with the wavelength 300 NM?
To answer you first question: First we need the energy of one photon. Assuming you are familiar with the two formulae E=hf and c=fl where: c=speed of light /m.s^-1 | h= Planck constant / Js l=wavelength /m | f=frequency / Hz E= Energy / J we can substitute to get E=hc/l l = 320nm = 320 x10^-9 m | E = 6.626 x10^-34 Js | c=3 x10^8 m.s^-1 so E=(6.626 x10^-34 x 3 x10^8)/(320 x10^-9) = 6.211875 x10^-19 [seems believable so far] One mole = 6.02 x10^23 So total energy of a mole of the photons is: 6.02 x10^23 x 6.211875 x10^-9 = 373954.875 J = 374 kJ (3 s.f.) --------- As for the second question: http://en.wikipedia.org/wiki/Ultraviolet#Subtypes there you can see the relative wavelenghts of different types of UV light. UVA = 400-320nm | UVB = 320-200nm Now, looking again at the equation E=hc/l It is clear that to maximise E, we want the smallest l (to divide hc by the smallest number) Therefore shorter wavelength photons have the highest energy, so the answer is UV-B
