When radiation of frequency $8 \times 10^{15} \ Hz$ is incident on a metal surface with a work function of $6.125 \ eV$,what is the kinetic energy of the emitted photoelectrons in $eV$?

$A$ metal surface of work function $1.13 eV$ is irradiated with light of wavelength $310 nm$. The retarding potential required to stop the escape of photoelectrons is [Take $hc = 1240 eV \cdot nm$] (in $V$)

If the work function of a photometal is $6.825 \; eV$,its threshold wavelength will be ............. $\mathring A$ $(c = 3 \times 10^8 \; m/s)$.

$A$ beam of light of wavelength $\lambda$ falls on a metal having work function $\phi$ placed in a magnetic field $B$. The most energetic electrons, moving perpendicular to the field, are bent in circular arcs of radius $R$. If the experiment is performed for different values of $\lambda$, then the $B^2$ vs. $\frac{1}{\lambda}$ graph will look like (keeping all other quantities constant):

Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 \, nm$. If the stopping potential for the ejected electrons is $2.8 \, eV$, then the work function of the metal is (Take $hc = 1240 \, eV \cdot nm$). (in $ \, eV$)

Photons with energy $5\, eV$ are incident on a cathode $C$ of a photoelectric cell. The maximum kinetic energy of the emitted photoelectrons is $2\, eV$. When photons of energy $6\, eV$ are incident on $C$,no photoelectrons will reach the anode $A$ if the stopping potential of $A$ relative to $C$ is .............. $V$.