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What is the energy of a quantum of light of frequency 6.31x10^14?

User FrankM
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Final answer:

The energy of a quantum of light, or photon, with a frequency of 6.31x10^14 Hz is found using the formula E = hv. Using Planck's constant, the calculated energy is 4.183 x 10^-19 Joules.

Step-by-step explanation:

To find the energy of a quantum of light (photon) with a frequency of 6.31×1014 Hz, we use the formula E = hv, where h is Planck's constant and v is the frequency of the light. Planck's constant h has a value of 6.626 × 10-34 J·s. By substituting the values into the formula, we get E = (6.626 × 10-34 J·s) × (6.31×1014 Hz), which gives us the energy of the photon.

Performing the multiplication, the energy of the photon is E = 4.183 × 10-19 Joules.

User Octavius
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