Light of frequency $4\nu_0$ is incident on a metal surface with a threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is:

The work functions for metals $A, B$,and $C$ are $1.92 \ eV, 2.0 \ eV$,and $5 \ eV$ respectively. According to Einstein's photoelectric equation,which metal$(s)$ will emit photoelectrons when irradiated with light of wavelength $4100 \ \mathring A$?

According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

$A$ 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 $\,s$)

In the graph given below,if the slope is $4.12 \times 10^{-15} \ V-sec$,then the value of $h$ should be:

For a photoelectric cell,the graph showing the variation of stopping potential $(V_o)$ with frequency $(\nu)$ of incident light is best represented by: