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-function of a metal is $1 eV$. Light of wavelength $3000 \text{Å}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photo-electrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then:

In an experiment on photoelectric emission from a metallic surface,the wavelength of incident light is $2 \times 10^{-7} \,m$ and the stopping potential is $2.5 \,V$. The threshold frequency of the metal (in $Hz$) is approximately (charge of electron $e=1.6 \times 10^{-19} \,C$,Planck's constant $h=6.6 \times 10^{-34} \,J-s$):

$A$ photon of energy $3.4 \ eV$ is incident on a metal having work function $2 \ eV$. The maximum kinetic energy $(K.E.)$ of photo-electrons is equal to ........... $eV$. (in $eV$)

How does the maximum kinetic energy of a photoelectron depend on the frequency of incident radiation? On which factors does it not depend?