The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface,having work function $5.01 \, eV$,when ultraviolet light of $200 \, nm$ falls on it,must be ............... $V$.

In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

$A$ metal surface is illuminated by radiation of wavelength $4500 \; \mathring A$. The ejected photoelectron enters a constant magnetic field of $2 \; mT$ making an angle of $90^{\circ}$ with the magnetic field. If it starts revolving in a circular path of radius $2 \; mm$,the work function of the metal is approximately ............. $eV$.

Statement $1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of photoelectrons is $K_{max}$. When $X$-rays are used instead of ultraviolet light, both $V_0$ and $K_{max}$ increase.
Statement $2$: Photoelectrons are emitted with a range of speeds from $0$ to a maximum value because the incident light contains a range of frequencies.

The stopping potential $V_0$ (in $volt$) as a function of frequency $(\nu)$ for a sodium emitter is shown in the figure. The work function of sodium,from the data plotted in the figure,will be: ................. $eV$
(Given: Planck's constant $(h) = 6.63 \times 10^{-34} \, Js$,electron charge $e = 1.6 \times 10^{-19} \, C$)

$A$ photon of energy $5.5 \ eV$ strikes a surface that emits photoelectrons with a maximum kinetic energy of $4.0 \ eV$. The stopping potential for these electrons is ............ $V$.