Question

Bulb of ${{40W}}$ is producing a light of wavelength ${{620 nm}}$ with ${{80\% }}$ of efficiency, then the number of photons emitted by the bulb in ${{20}}$ seconds are:
(${{1eV = 1}}{{.6 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$,${{hc = 12400 eV}}$)
1.$2 \times {10^{18}}$
2.${10^{18}}$
3.${10^{21}}$
4.$2 \times {10^{21}}$

Answer 1

Watt is equal to the work (joule) per second. First, we have to calculate the energy emitted by the ${{100\% }}$ efficiency bulb by using power formula and then calculate the ${{80\% }}$ efficiency bulb. Power of the bulb can be calculating the ratio of energy produced by bulb and time.

Complete step by step answer:
Power of a bulb =${{40 W}}$
Wavelength ${{\lambda = 620 nm}}$
${{1nm = 1}}{{{0}}^{{{ - 9}}}}$
${{620 nm = 620 }} \times {{1}}{{{0}}^{{{ - 9}}}}$
Let’s calculate the energy emitted by a bulb;(assume bulb efficiency is ${{100\% }}$)
${{Power = }}\dfrac{{{{Energy}}}}{{{{Time}}}}$………………………………….(1)
Rearrange the equation (1)
${{Energy = Power \times time}}$
= $40 \times 20 = 800J$
Hence, the energy emitted by the ${{100\% }}$ efficiency bulb is ${{800J}}$.
And the given bulb is ${{80\% }}$ efficiency.
Let’s calculate the energy emitted by a bulb which has ${{80\% }}$ efficiency.
${{Energy = }}\left[ {\dfrac{{{{80}}}}{{{{100}}}}} \right]{{ \times 800 = 640J}}$
Hence, energy emitted by the ${{80\% }}$ efficiency bulb is ${{640J}}$.
Let’s calculate the energy of a photon.
${{Energy of a photon = }}\dfrac{{{{hc}}}}{{{\lambda }}}$…………………………..(2)
${{h}}$= Planck's constant
${{c}}$= Velocity of light
${{\lambda }}$ = Frequency of light
From the given,
${{hc = 12400 eV}}$
wavelength = ${{620 nm}}$
${{1nm = 10}}{{{A}}^{{o}}}$
${{620 nm = 6200}}{{{A}}^{{o}}}$
Substitute the all given values in equation (2)
${{ = }}\dfrac{{{{12400 ev }}{{{A}}^{{o}}}}}{{{{6200 }}{{{A}}^{{o}}}}}{{ = 2ev}}$
Hence, the energy of the photon is ${{2ev}}$.
Let’s calculate the number of photons.
${\text{Number of photon = }}\dfrac{{{\text{Energy of light}}}}{{{\text{Energy of photon}}}}$
Energy of ${{80\% }}$ efficiency light = ${{640J}}$
${{1eV = 1}}{{.6 \times 1}}{{{0}}^{{{ - 19}}}}{{J}}$
${{ = }}\dfrac{{{{640J}}}}{{{{2eV}}}}{{ = }}\dfrac{{{{640 J}}}}{{{{2 \times 1}}{{.6 \times 1}}{{{0}}^{{{ - 19}}}}}}{{ = 2 \times 1}}{{{0}}^{{{21}}}}$
Hence, the correct answer is (4).

Additional information:
The photon is the electromagnetic wave carrier responsible for light, radio waves, microwaves, X-rays, and so on. All these waveforms are based on the same electromagnetic wave, but their wavelengths distinguish them. The wavelength of photons has an opposite relationship to the energy of photons.

Note: An electron collides with each particle of light, called a photon, and uses some of its energy to eject the electron. The remaining energy of the photon is transferred to a free negative charge, called a photoelectron. Photoelectrons are not emitted in the visible light because they have low threshold frequency energy.