In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

Monochromatic radiation of wavelength $640.2 \;nm$ $(1 \;nm = 10^{-9} \;m)$ from a neon lamp irradiates a photosensitive material made of caesium on tungsten. The stopping voltage is measured to be $0.54 \;V$. The source is replaced by an iron source and its $427.2 \;nm$ line irradiates the same photo-cell. Predict the new stopping voltage (in $V$). (in $;V$)

The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?

In an experiment to study the photoelectric effect,the observed variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident radiation is as shown in the figure. The slope and $y$-intercept are respectively

When a photosensitive surface is irradiated by light of wavelengths $\lambda_{1}$ and $\lambda_{2}$,the kinetic energies of the emitted photoelectrons are $E_{1}$ and $E_{2}$ respectively. The work function of the photosensitive surface is

Which of the following graphs represents the variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident light for a photoelectric cell?