The ratio of work functions of two metals $A$ and $B$ is $1:2$. If radiations of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of the emitted photoelectrons will be: (Given $f >$ threshold frequency of $A$ and $2f >$ threshold frequency of $B$)

When photons of energy $5 eV$ are incident on a metal surface with a work function of $4.36 eV$,photoelectrons are emitted. What is the maximum momentum of the emitted photoelectrons?

$A$ point source of light is used in a photoelectric effect. If the source is moved farther from the emitting metal, the stopping potential:

In a photoelectric emission experiment,the stopping potential for a given metal is $V$ volt,when radiation of wavelength $\lambda$ is used. If radiation of wavelength $2 \lambda$ is used with the same metal,then the stopping potential (in volt) will be. [Given: $c = \text{velocity of light}$,$e = \text{charge on electron}$,$h = \text{Planck's constant}$]

When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1} > W$), then the photocurrent falls to zero at a stopping potential of $V_{1}$. If the frequency of light is increased to $v_{2}$, the stopping potential changes to $V_{2}$. Therefore, the charge of an electron $e$ is given by:

Two radiations of photon energies $1 \; eV$ and $2.5 \; eV$ successively illuminate a photosensitive metallic surface of work function $0.5 \; eV$. The ratio of the maximum speeds of the emitted electrons is