In photoelectric effect, the K.E. of electrons emitted from the metal surface depends upon
In photoelectric effect, the K.E. of electrons emitted from the metal surface depends upon
- A
Intensity of light
- B
Frequency of incident light
- C
Velocity of incident light
- D
Both intensity and velocity of light
Similar Questions
An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3\, {eV}$. The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength $4000\, \mathring {{A}}$. What is the maximum kinetic energy of the emitted photoelectron ? (In ${eV}$)
An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3\, {eV}$. The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength $4000\, \mathring {{A}}$. What is the maximum kinetic energy of the emitted photoelectron ? (In ${eV}$)
- [JEE MAIN 2021]
A photon of wavelength $4400 \, \mathring A$ is passing through vacuum. The effective mass and momentum of the photon are respectively
A $200\, W$ sodium street lamp emits yellow light of wavelength $0.6\, \mu \,m$. Assuming it to be $50\%$ efficient in converting electrical energy to light, the number of photon of yellow light it emits per second is
A $200\, W$ sodium street lamp emits yellow light of wavelength $0.6\, \mu \,m$. Assuming it to be $50\%$ efficient in converting electrical energy to light, the number of photon of yellow light it emits per second is
A source $S_1$ is producing, $10^{15}$ photons per second of wavelength $5000 \;\mathring A.$ Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100\;\mathring A$. Then, $($ power of $S_2)/$ $($ power of $S_1)$ is equal to
A source $S_1$ is producing, $10^{15}$ photons per second of wavelength $5000 \;\mathring A.$ Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100\;\mathring A$. Then, $($ power of $S_2)/$ $($ power of $S_1)$ is equal to
- [AIPMT 2010]
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
$(a)$ The number of photons emitted per second by a Medium wave transmitter of $10\; kW$ power, emitting radiowaves of wavelength $500\; m$.
$(b)$ The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can percetve $(-10^{-10}\; W m ^{-2}$). Take the area of the pupil to be about $0.4 \;cm ^{2}$, and the average frequency of white light to be about $6 \times 10^{14}\; Hz$
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
$(a)$ The number of photons emitted per second by a Medium wave transmitter of $10\; kW$ power, emitting radiowaves of wavelength $500\; m$.
$(b)$ The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can percetve $(-10^{-10}\; W m ^{-2}$). Take the area of the pupil to be about $0.4 \;cm ^{2}$, and the average frequency of white light to be about $6 \times 10^{14}\; Hz$