When radiation of wavelength $400 \ nm$ is incident on a metal surface,the kinetic energy of the emitted photoelectrons is $1.68 \ eV$. The work function of the metal is .......... $eV$.

The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

If a photocell is illuminated with a radiation of $1240 \mathring{A}$,the stopping potential is found to be $8 \text{ V}$. Then the work function of the emitter and the threshold wavelength are:

In the experiment of $P.E.E.$,the $KE_{max}$ of an electron is $K_0$. If the frequency is increased by a factor of $n_1$,then the $KE_{max}$ becomes $n_2K_0$. Find the work function.

Given below are two statements:
Statement $I$: Stopping potential in photoelectric effect does not depend on the power of the light source.
Statement $II$: For a given metal,the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.
In the light of the above statements,choose the most appropriate answer from the options given below.

In an experiment on photoelectric emission from a metallic surface,the wavelength of incident light is $2 \times 10^{-7} \,m$ and the stopping potential is $2.5 \,V$. The threshold frequency of the metal (in $Hz$) is approximately (charge of electron $e=1.6 \times 10^{-19} \,C$,Planck's constant $h=6.6 \times 10^{-34} \,J-s$):