The detector completely absorbs light incident on it with a frequency of v = 5 • 10 ^ 14 Hz.

The detector completely absorbs light incident on it with a frequency of v = 5 • 10 ^ 14 Hz. The absorbed power is equal to P = 3.3 • 10 ^ -14 W. How many photons are incident on the detector during t = 5 s? Planck’s constant 6.6 * 10 ^ -34 J * s.

To find the number of photons falling on the taken detector, we will use the formula: P = Eph * N / t = h * ν * N / t, whence we express: N = P * t / (h * ν).

Constants and variables: P – power absorbed by the detector (P = 3.3 * 10 ^ -14 W); t – elapsed time (t = 5 s); h – Planck’s constant (by condition h = 6.6 * 10 ^ -34 J * s); ν is the frequency of the light incident on the detector (ν = 5 * 10 ^ 14 Hz).

Let’s calculate: N = P * t / (h * ν) = 3.3 * 10 ^ -14 * 5 / (6.6 * 10 ^ -34 * 5 * 10 ^ 14) = 5 * 10 ^ 5 photons.

Answer: 5 * 10 ^ 5 photons will fall on the taken detector in 5 seconds.