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

Light of wavelength $4000 \,\mathring A$ falls on a photosensitive metal and a negative $2 \,V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 \times 10^{-34} \,Js, \,e = 1.6 \times 10^{-19} \,C, \,c = 3 \times 10^8 \,m/s)$.

In case of photoelectric emission from a certain metal, the cutoff frequency is $\nu$. If radiation of frequency $3\nu$ is incident on the metal plate, the maximum possible velocity of the emitted electrons will be ($m$ = mass of electron, $h$ = Planck's constant).

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 $4100 \ \mathring A$?

The maximum velocity of photoelectrons emitted from a metal surface is $1.2 \times 10^6 \ m/s$. If the specific charge of the electron is $1.8 \times 10^{11} \ C/kg$,then the value of the stopping potential in volts is:

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.