In a photoelectric effect experiment, for radiation with frequency `v_0` with `hv_0` = 8eV, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 `v_0`?

In a photoelectric effect experiment, the maximum kinetic energy of the emitted electrons is 1eV for incoming radiations of frequency v_(0) and 3eV for incoming radiation of frequency 3v_(0)//2 . What is the maximum kinetic energy of the electrons emitted for incomming radiation of frequency 9v_(0)//4 ?

In a triode, cathode, grid and plate are at 0, – 2 and 80 V respectively. The electrons is emitted from the cathode with energy 3 eV . The energy of the electron reaching the plate is

In the explanation of photoelectric effect, we assume one photon of frequency v collidies with an electron and transfers its energy. This leads to the equation for the maximum energy E_max of the emitted electron as E_max=hv-phi_0 Where phi_0 is the work function of the metal. if an electron absorbs 2 photons (each of frequency v) what will be the maximum energy for the emitted electron?

The threshold frequency V_0 for a metal is 7.0xx10^14 s^1 . Calculate the kinetic energy of an electron emitted when radiation of frequency v=1.0xx10^15s^(-1) hits the metal.

(i) In the explanation of photoelectric effect we assume one photon of frequency with v collides with an electron and transfers its energy.This lead to the equation for the maximum energy E_(max) of the emitted electron as E_(max)=hv-phi_(0) Where phi_(0) is the work function of the metal If an electron absorbs 2 photon (each of frequency v)What will be the maximum energy for the emitted electron? (ii)Why is this fact (two photon absorption)not taken into consueration in our discussion of the stopping potential?

(i) In the explanations of photoelectric effect, we assume one photon of frequency v collides with an electron and transfer its energy. This leads to the equation for the maximum energy E_(max) of the emitted electron as E_(max)=hv-phi_(0) Where phi_(0) is the work function of the metal. if an electron absorbs 2 photons (each of frequency v) what will be the maximum energy for the emitted electron? (ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

When a high frequency electromagnetic radiation is incident on a metallic surface, electrons are emitted from the surface. Energy of emitted photoelectrons depends only on the frequency of incident electromagnetic radiation and the number of emitted electrons depends only on the intensity of incident light. Einstein's protoelectron equation [K_(max)=hv-phi] correctly ecplains the PE, where upsilon= frequency of incident light and phi= work function. Q. For photoelectric effect in a metal, the graph of the stopping potential V_0 (in volt) versus frequency upsilon (in hertz) of the incident radiation is shown in Fig. The work function of the metal (in eV) is.

When a high frequency electromagnetic radiation is incident on a metallic surface, electrons are emitted from the surface. Energy of emitted photoelectrons depends only on the frequency of incident electromagnetic radiation and the number of emitted electrons depends only on the intensity of incident light. Einstein's protoelectron equation [K_(max)=hv-phi] correctly ecplains the PE, where upsilon= frequency of incident light and phi= work function. Q. For photoelectric effect in a metal, the graph of the stopping potential V_0 (in volt) versus frequency upsilon (in hertz) of the incident radiation is shown in Fig. The work function of the metal (in eV) is.

A metal having a threshold frequency of v_0//2 is illuminated by the incident radiation of frequency 2v_0 . If the energy of the incident photon is increased by 40 percent , then what will be the percentage change in the kinetic energy of the photoelectrons emitted from the surface of the metal ? Note that increase in energy of photon is done by suitably changing frequency of radiation.