Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminate a metallic surface whose work function is $0.5 \text{ eV}$. The ratio of the maximum speeds of the emitted electrons will be:

In a photoelectric effect experiment,the maximum kinetic energy of emitted photoelectrons is $K_0$. If the frequency of the incident radiation is increased by a factor of $n_1$,the new maximum kinetic energy becomes $n_2K_0$. Find the work function of the metal.

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^{6} \ m/s$. Take the value of specific charge of the electron as $1.8 \times 10^{11} \ C/kg$. Then the stopping potential in volt is

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630 \; \mathring{A}$ is $0.42 \; V$. If the threshold frequency is $x \times 10^{13} \; s^{-1}$,where $x$ is ... (nearest integer).
(Given: speed of light $= 3 \times 10^{8} \; m/s$,Planck's constant $= 6.63 \times 10^{-34} \; J \cdot s$)

The maximum velocity of electrons emitted from a metal surface is $V$,when the frequency of light falling on it is $f$. What is the maximum velocity when the frequency becomes $4f$?

$A$ photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$,then the radius $r$ is given by,(in the usual notation)