The photoelectric work function for a metal is $2.4 \ eV$. Among the four wavelengths,the wavelength of light for which photoemission does not take place is: (in $nm$)

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

On a photosensitive material,when the frequency of incident radiation is increased by $30 \%$,the kinetic energy of emitted photoelectrons increases from $0.4 \ eV$ to $0.9 \ eV$. The work function of the surface is (in $eV$)

The maximum velocity of the photoelectron emitted by the metal surface is $v$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volt is

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy, for the same wavelength of incident radiation?

The work function of a metal is $1 \; eV$. Light of wavelength $3000 \; \mathring{A}$ is incident on this metal surface. The velocity of the emitted photo-electrons will be: