Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

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 .............. $V$.

In a photoelectric experiment,electrons are ejected from metals $X$ and $Y$ by light of intensity $I$ and frequency $f$. The potential difference $V$ required to stop the electrons is measured for various frequencies. If $Y$ has a greater work function than $X$,which one of the following graphs best illustrates the expected results?

When radiations of wavelength $\lambda$ are incident on a metallic surface,the stopping potential required is $4.8 \ V$. If the same surface is illuminated with radiations of double the wavelength,the required stopping potential becomes $1.6 \ V$. The value of the threshold wavelength for the surface is:

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1: n$,then the threshold frequency of the metallic surface is

If the threshold wavelength for sodium is $6800 \mathring{A}$,then the work function will be ............. $eV$.