In the photoelectric emission process from a metal with a work function of $1.8\, eV$,the kinetic energy of the most energetic electrons is $0.5\, eV$. The corresponding stopping potential is ......... $V$.

Assertion : Photosensitivity of a metal is high if its work function is small.
Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $(\lambda/3)$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $4$ times that in the first case, the work function of the surface of the material is ($h$ = Planck's constant, $c$ = speed of light).

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 ............... $V$.

The work function of a photoelectric material is $3.3 \text{ eV}$. The threshold frequency will be equal to

In a photoelectric emission experiment,the stopping potential for a given metal is $V$ volt,when radiation of wavelength $\lambda$ is used. If radiation of wavelength $2 \lambda$ is used with the same metal,then the stopping potential (in volt) will be. [Given: $c = \text{velocity of light}$,$e = \text{charge on electron}$,$h = \text{Planck's constant}$]