Light of wavelength `3500Å` is incident on two metals A and B. Which metal will yield photoelectrons if their work functions are 4.2 eV and 1.9 eV respectivley ?

Light of wavelength 3500Å is incident on two metals A and B whose work functions are 3.2 eV and 1.9 eV respectively. Which metal will emit photoelectrons

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. Light of wavlength 3300 is incident on two metals A and B, whose work function are 4 eV and 2 eV, respectively. Then

Light of wavelength 4000 Å is incident on a metal surface. The maximum kinetic energy of emitted photoelectron is 2 eV. What is the work function of the metal surface ?

Light of waelength 4000Å is incident on a metal plate with work function 2eV. The max. K.E. of the photoelectrons is

Light of wavelength 4000 Å is incident on a sodium surface for which the threshold wavelength of photoelectrons is 5420 Å. The work function of sodium is

Light of wavelength 3000Å is incident on a metal surface whose work function is 1 eV. The maximum velocity of emitted photoelectron will be

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV . The de Broglie wavelength of the emitted electron is

When light of wavelength lesser than 6000 Å is incident on a metal, electrons are emitted. The approximate work - function of the metal is :

Ligth of wavelength 4000 Å is incident on a metal plate whose work function is 2 eV . What is maximum kinetic enegy of emitted photoelectron ?