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

A 100 W incandescent bulb converts approximately 2.5% of electrical energy into visible light. The energy of each photon at an average wavelength of 540 nm in electronvolts is calculated using the energy formula E = hc/λ. Subsequently, based on the bulb's light output and light geometry, the number of visible photons entering an aperture is determined.

Step-by-step explanation:

The student's question involves calculating the energy of photons emitted by a light bulb, and determining how many photons would enter a specific area some distance away from it. We first need to calculate the energy of each photon in electronvolts (eV) before we can address the other parts of the question.

a) To find the energy of each photon, we use the formula E = hc/λ, where 'h' is Planck's constant (6.62607015 × 10⁻³⁴ Js), 'c' is the speed of light (3.00 × 10⁸ m/s), and 'λ' is the wavelength of the light in meters (540 nm = 540 × 10⁻⁹ m). However, to convert this energy to eV, we must use the conversion factor: 1 eV = 1.602 × 10⁻ J.

b) To find the number of photons per second, N, we need to determine the total power output in visible light, which is 2.5% of 100 W. Then we divide this by the energy per photon, accounting for the geometry of light spread and the given aperture area (A = 2 cm²). Note, the solid angle subtended by the aperture and the distance from the bulb (D = 5 m) will influence the number of photons.

c) If the average wavelength changes to 810 nm, the energy per photon will change, which affects the number of photons per second that would enter the same aperture.

Answer 2

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.