Question

A 100 W incandescent light bulb converts approximately 2.5% of the electrical energy supplied to it into visible light. Assume that the average wavelength of the emitted light is λ = 540 nm, and that the light is radiated uniformily in all directions.

a) What is the energy of each photon in eV?

b)How many photons per second, N, would enter an aperture of area A = 2cm2 located a distance D = 5 m from the light bulb?

c)Suppose, instead, that the average photon wavelength is 810 nm. How many photons per second, N', would enter the aperture?

Answer 1

Answer: a) E photon= 2.39 eV;  b) 6.32* 10^48 fotones/s c) 9,87 * 10^48 fotones/s

Explanation: In order to calculate the quenstion we have to consider the Plack postulate that the energy of one photon is given by

Ephoton=  h*f where h is the Planck universal constant and f the frequency of ligth

so, we also known that for waves the frequency and wavelength are related  in the form:

velocity= wavelength (λ)*frequency (f) in our case velocity is equal to the speed of light, thus we can obtain

Ephoton= h*f= h*c/λ=   also hc= 1240 eV/nm

so Ephoton=1240/540 nm= 2.39 eV

The power is energy/time so if we considerer a number per second (n) multiplier by its energy we obtain the power emited

in this case Power: n*h*f so we have n= power/h*f

On the other hand to calculate the number of photons per second at 5 m from light bulb considering an area of 2 cm^2, we have to consider that the light is emitted isotropically  so

the total area emitted at 5m is 4π25, the fraction in the 2 cm^2 is

0.02*0.02/25=1,6*10^-5

For photons for 810 nm we have to change photon energy to 1.53 eV to carry out the calculations.

As 810 nm photons have lower energies for light bulb emitting at 100 W give more that these photons.