Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.

Photons of wavelength $\lambda$ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius $R$ by a perpendicular magnetic field having a magnitude $B$. The work function of the metal is (where,symbols have their usual meanings)

$UV$ light of $4.13 eV$ is incident on a photosensitive metal surface having work function $3.13 eV$. The maximum kinetic energy of ejected photoelectrons will be : (in $eV$)

Which one of the four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident radiation $(I)$ is correct?

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $3V_0$. When the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

In a photoelectric cell,the wavelength of incident light is changed from $4000 \, \mathring{A}$ to $3600 \, \mathring{A}$. The change in stopping potential will be ............. $V$.