The energy of one photon of light with a frequency of 7.03 × 1014 Hz is approximately 4.65618 × 10-19 Joules. This is obtained using Planck's equation, E = hf, where E is the energy, h is Planck's constant and f is the frequency of the light.
Explanation:
The subject of this question is Physics. The student wants to know the energy of exactly one photon of light that has a frequency of 7.03 × 1014 Hz. We can use Planck's equation which relates the energy of photon, its frequency and Planck's constant (h = 6.626 × 10-34 J.s).
So, the energy (E) of the photon is given by the formula E = hf. Substituting the frequency (f = 7.03 × 1014 Hz) and Planck's constant into the formula gives E = (6.626 × 10-34 J.s)(7.03 × 1014 Hz) which gives approximately 4.65618 × 10-19 Joules. This is the energy of one photon of light with a frequency of 7.03 × 1014 Hz.
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