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A dark-adapted human eye at the peak of its sensitivity (510 nm) can perceive a flash when 3.5 x 1015 J of energy enter the iris. How many photons is this? (In fact only 10% of these are absorbed by the retina.)

User Mrco
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1 Answer

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Answer:

In 3.5 x 10^(15) J of energy there are 9*10^(33) photons.

Step-by-step explanation:

To solve this problem, we need two equations.

The equation of light velocity, wich is a relation between wavelenght and frecuency.

c=λν (1)

where:

  • c: speed of light = 3 × 10^8 [m/s]
  • ν: frecuency [1/s]
  • λ: wavelenght of wave [m]

The Photoelectric Effect equation, that refers to the energy absorbed or emanate by ONE photon.

E = hν (2)

where:

  • h : Planck´s constant = 6,626*10^{-34} [J.s]
  • ν: frecuency of radiation [s]
  • Ef: energy of one photon [J]

The first we do is to calculate the frecuency of the flash using equation (1). The wavelenght of the flash is 510 nm = 510 * 10^(-9) m

c=λν........................ ν= c/λ = 3 × 10^8 [m/s]/ 510 * 10^(-9) m = 5,88 * 10^(14) 1/s

Note: small wavelenghts always have big frequencies

Now, we use the photoelectric effect equation to calculate the amount of energy that ONE photon can abosrb.

E = hν ..................... E = 6,626*10^{-34} [J.s] * 5,88 * 10^(14) 1/s =3,9 * 10^(-19) J

To know the number of photons, we just have to divide the TOTAL amount of energy between the energy of ONE photon. So:

# photons = 3.5 x 10^(15) J / 3,9 * 10^(-19) J = 9*10^(33) photons.