The work functions for metals $A, B$ and $C$ are $1.92 eV, 2.0 eV$ and $5 eV$ respectively. According to Einstein's photoelectric equation,which of these metals will emit photoelectrons for an incident radiation of wavelength $4100 \mathring{A}$?

The work functions of metals $A$ and $B$ are in the ratio $1 : 2$. If light of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energy of photoelectrons emitted will be ($f$ and $2f$ are both frequencies greater than the threshold frequency of metals $A$ and $B$).

The work function of sodium is $2.3 eV$. The threshold wavelength of sodium will be ........... $\mathring A $

For zero photoelectric current,the stopping potential is:

$A$ monochromatic point source of light is placed at a distance $d$ from a metal surface. Photoelectrons are ejected at a rate $n$ per second,and with maximum kinetic energy $E$. If the source is brought nearer to a distance $d / 2$,the rate and the maximum kinetic energy per photoelectron become nearly

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is:
Where $h =$ Planck's constant, $m =$ mass of electron, and $c =$ speed of light.