If the maximum velocity with which an electron can be emitted from a photocell is $4 \times 10^8 \, cm/s$,the stopping potential is ................ $V$ (mass of electron $= 9 \times 10^{-31} \, kg$).

The figure shows different graphs between stopping potential $(V_0)$ and frequency $(\nu)$ for photosensitive surfaces of cesium,potassium,sodium,and lithium. The plots are parallel. The correct ranking of the targets according to their work function,with the greatest first,is:

Radiation of wavelength $332 \ nm$ is incident on a metal surface having a work function of $1.07 \ eV$. The stopping potential required to stop the emission of photoelectrons from the metal surface is ............ $V$. $(h = 6.6 \times 10^{-34} \ J s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

$A$ light source is placed at a distance of $20 \ cm$ from a photocell,and the measured stopping potential is $0.6 \ V$. If the distance of the source is changed to $40 \ cm$,what will be the stopping potential in $V$?

$A$ light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2 \ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of the metal is $1 \ \text{eV}$). (in $\text{eV}$)

When light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are just liberated. For another emitter,light of wavelength $600 \ nm$ is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is