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A microwave photon has a wavelength of 21 cm What is its energy?
Wiki User
∙ 14y ago
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]
Wiki User
∙ 14y ago
Wiki User
∙ 6y agoMicrowaves have a wavelength of 12.2 cm (abbreviated with a lambda) and a frequency (f) of 2.45 GHz. Einstein discovered that the magnitude of an energy quantum (a microwave photon in this case) is proportional 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 J*s), and f is the frequency. Plugging in the numbers we get
E = (6.626*10^-34 J*s) * (2,450,000,000 Hz) = 1.62337*10^-24 J
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