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The energy of the electromagnetic wave can be calculated using the formula E = hf, where h is Planck's constant and f is the frequency. Plugging in the values, E = (6.626 x 10^-34 J s) * (5.0 x 10^5 Hz) = 3.313 x 10^-28 J.

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6.6262 x 1034 Js is?

The value 6.6262 x 10^34 Js represents the Planck constant, denoted as h, in joules seconds. It is a fundamental constant in quantum mechanics and is used to describe the relationship between energy and frequency of electromagnetic radiation.


What is 33 percent times 1034?

33% * 1034= 0.33 * 1034= 341.22


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1034


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1034 * 7 = 7,238


What is 12 times 1034?

12 x 1034 = 12408


What is 2080 divided by 1034?

2.0116


What is the prime factorization of 1034?

2 x 11 x 47 = 1034


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1 + 2 + 11 + 22 + 47 + 94 + 517 + 1034 = 1,728


What are the factors of 1034?

1,2,11,22,47,94,517,1034


What frequency of light of light has energy of 8.2x1019 j?

When you talk about the energy of any electromagnetic radiation in terms of itsfrequency, you're talking about the energy of a single photon.8.2 x 1019 J is a bit more than 1,000 times the energy that the Braidwood nucleargenerating station south of Chicago produces in a year.In order for a single photon to have 8.2 x 1019 J of energy, its frequency would have to be8.2 x 1019/Planck's Konstant = 8.2 x 1019/6.62608 x 10-34 = 1.2375 x 1053 Hz.That's about 1034 times the frequency a photon needs in order to be called agamma-ray. At that frequency, the wavelength is about 2.422 x 10-45 meter,and that's something like 10-27 the size of an electron.Perhaps you meant to type 8.2 x 10minus 19 J.The frequency of photon with that energy is8.2 x 10-19/6.62608 x 10-34 = 1.238 x 1015 Hz.and its wavelength is about 242 nanometers.That would be a photon in the mid-range ultraviolet.If you want a beam of light that carries your alleged 8.2 x 10plus 19 J,you just need more of these photons ... like 1038 of them.


What are the first 10 multiples of 1034?

1034, 2068, 3102, 4136, 5170, 6204, 7238, 8272, 9306, 10340.


An sf 1034 is a what document?

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