If in a photoelectric experiment,the wavelength of incident radiation is reduced from $6000 \, \mathring{A}$ to $4000 \, \mathring{A}$,then:

$A$ metal surface having work function '$W_{0}$' emits photoelectrons when photons of energy '$E$' are incident on it. The electron enters a uniform magnetic field '$B$' in a perpendicular direction and moves in a circular path of radius '$r$'. Then '$r$' is equal to (where '$m$' and '$e$' are the mass and charge of the electron,respectively).

Two photons of energy $2.5 eV$ and $3.5 eV$ fall on a metal surface of work function $1.5 eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

The beam of light has three wavelengths $4144 \,\mathring A$,$4972 \,\mathring A$,and $6216 \,\mathring A$ with a total intensity of $3.6 \times 10^{-5} \, W/m^2$ equally distributed amongst the three wavelengths. The beam falls normally on an area of $1 \, cm^2$ of a clean metallic surface with a work function of $2.3 \, eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2 \, s$.

The work function of a metal is $2 \ eV$. If a radiation of wavelength $3000 \ \text{Å}$ is incident on it,the maximum kinetic energy of the emitted photoelectrons is (Planck's constant $h=6.6 \times 10^{-34} \ \text{Js}$; velocity of light $c=3 \times 10^8 \ \text{m/s}$; $1 \ \text{eV}=1.6 \times 10^{-19} \ \text{J}$).

In a photoelectric effect experiment,if the frequency of light is doubled,the stopping potential will