The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is $1.65 \ eV$) will be

The threshold frequency for a metallic surface corresponds to an energy of $6.2 \ eV$ and the stopping potential for a radiation incident on this surface is $5 \ V$. The incident radiation lies in:

The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.

Assuming photoemission to take place,the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times,is

The surface of a metal is illuminated with light of $400 \ nm$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \ eV$. The work function of the metal is ............ $eV$ $(hc = 1240 \ eV \ nm)$.

The stopping potential $V_0$ (in $volt$) as a function of frequency $(\nu)$ for a sodium emitter is shown in the figure. The work function of sodium,from the data plotted in the figure,will be: ................. $eV$
(Given: Planck's constant $(h) = 6.63 \times 10^{-34} \, Js$,electron charge $e = 1.6 \times 10^{-19} \, C$)