Asodium lamp emits light at the power p = 70.0 w and at the wavelength λ = 600 nm, and the emission is uniformly in all directions.

(a) at what rate are photons emitted by the lamp?

(b) at what distance from the lamp will a totally absorbing screen absorb photons at the rate of 1.00 photon /cm2s?

(c) what is the rate per square meter at which photons are intercepted by a screen at a distance of 1.90 m from the lamp?

a)  # = 21,116 10¹⁹ photons , b)  R = 4,099 10⁷ m , c)  I = 4.66 10¹⁸ photons / m²

Explanation:

a) In this problem we need to know the energy of each photon and the total emitted power

Let's start with the Planck equation

     E = h f

     c =λ f

     E = h c / λ

     E = 6.63 10⁻³⁴ 3 10⁸/600 10⁻⁹

     E = 3,315 10⁻¹⁹ J

This is the energy of a photon

The definition of power is

       P = W / t

Work is equal to energy change

        P = E / t

       E = 70.0  1

        = 70 J

This is the total energy emitted

Let's use a rule of proportions to find the number of photons

       # = / E

       # = 70 / 3,315 10⁻¹⁹

       # = 21,116 10¹⁹ photons

b) The definition of intensity is

       I = P / A

       P = I A

We want an intensity  

        I = 1 photon / cm²

The area of a sphere is

        A = 4π r²

        P = I 4π R²

        R = √ P / 4π I

The power is equal to the number of photons per energy of each photon emitted per unit of time

      R = √ (# / 4π I)

Intensity is

      I = 1 photon / cm² (100 cm / 1m) 2 = 1 10⁴ photon / m²

     R = √ (21,116 10¹⁹ / 4π 10⁴)

     R = √ (16.80 10¹⁴)

     R = 4,099 10⁷ m

c) we calculate the intensity

      A = 4π R²

      A = 4π 1.9²

      A = 45.36 m²

       I = P / A

       I = 21,116 10¹⁹ / 45.36

       I = 4.66 10¹⁸ photons / m²

c) power lifting, will be your answer.

,

deku ❤