The work functions of Silver and Sodium are $4.6 \ eV$ and $2.3 \ eV$,respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is :

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when light of wavelength $\lambda_1$ and $\lambda_2$ respectively are incident on a metallic surface. If $\lambda_1 = 3\lambda_2$ then

For the photoelectric effect,the maximum kinetic energy $E_k$ of the emitted photoelectrons is plotted against the frequency $\nu$ of the incident photons as shown in the figure. The slope of the curve gives

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy, for the same wavelength of incident radiation?

$A$ cobalt $(Co)$ plate is placed at a distance of $1 \,m$ from a point source of power $1 \,W$. Assume a circular area of the plate of radius $r = 1 \,Å$ is exposed to the radiation and ejects photoelectrons. The light energy is considered to be spread uniformly and the work function of cobalt is $5 \,eV$. The minimum time the target should be exposed to the light source to eject a photoelectron (assuming no reflection losses) is: (in $\,s$)

The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is: