Advertisements
Advertisements
Question
A 150 W lamp emits light of the mean wavelength of 5500 Å. If the efficiency is 12%, find out the number of photons emitted by the lamp in one second.
Advertisements
Solution
P = 150 W
λ = 5500 × 10−10 m
Efficiency = 12%
N = `("P"λ)/"hc"`
= `(150 xx 5500 xx 10^-10)/(6.626 xx 10^-34 xx 3 xx 10^8)`
= `(825 xx 10^-7)/(19.878 xx 10^-26)`
= 41.5 × 1019
Number of photons emitted is with 12% efficiency = `41.5 xx 10^19 xx 12/100`
N = 4.98 × 1019
APPEARS IN
RELATED QUESTIONS
Find the (a) maximum frequency and (b) minimum wavelength of X-rays produced by 30 kV electrons.
How will the thermionic current vary if the filament current is increased?
A diode value is connected to a battery and a load resistance. The filament is heated, so that a constant current is obtained in the circuit. As the cathode continuously emits electrons, does it become more and more positively charged?
Answer the following question.
Define the term "Threshold frequency", in the context of photoelectric emission.
The wavelength λe of an electron and λp of a photon of same energy E are related by ______.
If a light of wavelength 330 nm is incident on a metal with work function 3.55 eV, the electrons are emitted. Then the wavelength of the wave associated with the emitted electron is (Take h = 6.6 × 10–34 Js)
Emission of electrons by the absorption of heat energy is called ____________ emission.
Define the work function of a metal. Give its unit.
Photoelectric emission is observed from a metallic surface for frequencies ν1 and ν2 of the incident light (ν1 > ν2). If the maximum value of kinetic energy of the photoelectrons emitted in the two cases are in the ration 1 : n then the threshold frequency of the metallic surface is ______.
Name the factors on which photoelectric emission from a surface depends.
