The photoelectric cut-off voltage in a certain experiment is $1.5 \text{ V}$. The maximum kinetic energy of photoelectrons emitted will be . . . . . . .

Light of wavelength $488 \;nm$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38 \;V$. Find the work function (in $eV$) of the material from which the emitter is made.

The electric field at the point associated with a light wave is given by $E = 200 [\sin(6 \times 10^{15} t) + \sin(9 \times 10^{15} t)] \, Vm^{-1}$. Given $h = 4.14 \times 10^{-15} \, eVs$. If this light falls on a metal surface having a work function of $2.50 \, eV$,the maximum kinetic energy of the photoelectrons will be ........... $eV$.

Photoelectric effect can be explained by

Ultraviolet light of wavelength $200 \ nm$ is incident on a freshly polished surface of iron. The work function of the surface is $4.71 \ eV$. What will be the stopping potential in $eV$? $(h = 6.626 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J, c = 3 \times 10^8 \ m/s)$

The following graphs show the variation of stopping potential $(V_s)$ corresponding to the frequency of incident radiation $(f)$ for a given metal. The correct variation is shown in graph ($f_0 =$ Threshold frequency):