$A$ metal surface of work function $1.07 eV$ is irradiated with light of wavelength $332 nm$. The retarding potential required to stop the escape of photo-electrons is .............. $V$.

If the frequency of light falling on a photosensitive material doubles, which of the following is true?

The work function of caesium metal is $2.14 \; eV$. When light of frequency $6 \times 10^{14} \; Hz$ is incident on the metal surface,photoemission of electrons occurs. What is the
$(a)$ maximum kinetic energy of the emitted electrons,
$(b)$ stopping potential,and
$(c)$ maximum speed of the emitted photoelectrons?

$A$ photosensitive surface made of cesium on tungsten is irradiated by monochromatic radiation of wavelength $640.2 \ nm$ $(1 \ nm = 10^{-9} \ m)$ from a neon bulb. The measured stopping potential is $0.54 \ V$. If the source is replaced by another source and the same photocell is irradiated by a line of $427.2 \ nm$,what will be the new stopping potential in $V$?

In a photoelectric experiment, the slope of the graph drawn between stopping potential $(V_s)$ along the $y$-axis and the frequency $(\nu)$ of incident radiation along the $x$-axis is (Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$)

$UV$ 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 $eV$)