The detector completely absorbs light falling on it with a frequency of v = 6 • 10 ^ 14 Hz. During the time t = 5 s, N = 3 • 10 ^ 5 photons fall on the detector. What is the power absorbed by the detector?
v = 6 * 10 ^ 14 Hz.
t = 5 s.
N = 3 * 10 ^ 5.
h = 6.6 * 10 ^ -34 J * s.
R – ?
The absorbed power of the detector P is the energy E that the detector absorbs per unit time t.
The detector power is determined by the formula: P = E / t.
When light is absorbed, photons are absorbed. The energy of one photon E1, according to Planck’s formula, is E1 = h * v.
Where h is Planck’s constant, v is the frequency of the incident light.
E = E1 * N = h * v * N.
The formula for determining the absorbed power of the detector will take the form: P = h * v * N / t.
P = 6.6 * 10 ^ -34 J * s * 6 * 10 ^ 14 Hz * 3 * 10 ^ 5/5 s = 23.76 * 10 ^ -15 W.
Answer: the absorbed power of the detector is: P = 23.76 * 10 ^ -15 W.
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