Light in incident on a metal plate whose work function is 2 eV. Electric field associated with light is given by `E=E_(0)sin(omegat-(2pi)/(5xx10^(7))x)`{ S.I unit} if energy of photom is given by `(12375)/(lamda("in"A))` eV then stopping potential is
Electric field associated with a light wave is given E= E_(0) sin [1.57 xx10^(15)t +6.28 xx10^(15) t] V/m. If this light incident on a surface of work function 2.0 eV the stopping potential will be -
What is the stopping potential, when a metal surface with work function 1.2 eV is illuminated with light of energy 3 eV ?
Light of waelength 4000Å is incident on a metal plate with work function 2eV. The max. K.E. of the photoelectrons is
What is the stopping potential when the metal with work function 0.6 eV is illuminated with the light of 2 eV ?
Light of energy 2.0 eV falls on a metal of work function 1.4 eV . The stopping potential is
The electric field in space is given by overset(rarr)E=E_(0)sin (omegat+6y-8z)hatn then the direction of propagation of light wave is
Light of 5,500 Å is incident on a metal of work function 2 eV . Calculate the (i) threshold wavelength (ii) maximum velocity of electrons emitted.
The electric field associated with a light wave is given by E= E_0 sin [(1.57x 10^7 m^(-1)(x-ct)]. Find the stopping potential when this light is used in an experiment on photoelectric affect with a metal having work - function 1.9 eV.
