Light quanta with energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV. Find the maximum impulse transmitted to the surface of the metal when each electron flies out.

Wavelengths belonging to Balmer series lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work function is 2.0 eV. Find ( in eV ) the maximum kinetic energy of the emitted photoelectrons. ("Take hc = 1242 eV nm.")

The energy of emitted photoelectrons from a metal is 0.9ev , The work function of the metal is 2.2eV . Then the energy of the incident photon is

A photon energy 3.4 eV is incident on a metal having work function 2 eV . The maximum K.E. of photoelectrons is equal to

Light of energy 2.0 eV falls on a metal of work function 1.4 eV . The stopping potential is

Two potons of energy 2.5eV each are incident on a metal plate whose work function is 4.0eV , then the number of emitted from the metal surface will be

Two photons, each of energy 2.5 eV are simultaneously incident on the metal surface. If the work function of the metal is 4.5 eV, then from the surface of metal

Radiation of a certain wavelength causes electrons with a maximum kinetic energy of 0.65 eV to be ejected from a metal surface whose work function is 2.65 eV. What will be the maximum kinetic energy (in eV) with which this same radiation ejects electrons from another metal whose work function is 2.15eV.