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 figure shows the variation of photocurrent with anode potential for four different radiations. Let $f_a$,$f_b$,$f_c$ and $f_d$ be the frequencies for the curves $a$,$b$,$c$ and $d$ respectively.

$A$ metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2 \lambda$,the stopping potential becomes $\frac{V_0}{4}$. The threshold wavelength for this metallic surface will be -

$A$ light of wavelength $310 \,nm$ is used in a photoelectric experiment. The metal electrode of work function $2.5 \,eV$ is used in the experiment. The stopping potential for the photoelectrons will be (assume $hc = 1240 \,eV-nm$): (in $V$)

Light comprising three wavelengths $310 \ nm$,$455 \ nm$,and $620 \ nm$ is incident on a surface separating two media at an angle of $45^o$. The refractive index for the $455 \ nm$ light is $\sqrt{2}$. If a metal plate with a work function of $1.2 \ eV$ is placed in the second medium,what is the maximum kinetic energy of the electrons emitted from the metallic plate in $eV$?

The work functions for metals $A, B$,and $C$ are $1.92 \ eV, 2.0 \ eV$,and $5 \ eV$ respectively. According to Einstein's photoelectric equation,which metal$(s)$ will emit photoelectrons when irradiated with light of wavelength $4100 \ \mathring A$?