The energy of a photon with wavelength $\lambda$ is $2 \ eV$. When it strikes a metal surface,the maximum velocity of the emitted photoelectrons is $v$. If the value of $\lambda$ is decreased by $25\%$ and the maximum velocity is doubled,the work function of the metal becomes ...... $eV$.

The slope of the graph between stopping potential and the frequency of incident radiation for the photoelectric effect is . . . . . . .

The work function of $Zn$ is $4.25 \ eV$. The threshold frequency corresponds to which region of the electromagnetic spectrum?

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

Two radiations of photon energies $1 \; eV$ and $2.5 \; eV$ successively illuminate a photosensitive metallic surface of work function $0.5 \; eV$. The ratio of the maximum speeds of the emitted electrons is

$A$ monochromatic light of wavelength $\lambda$ ejects photoelectrons from a metal surface with work function $\phi = 2.4 \text{ eV}$. These photoelectrons are made to collide with hydrogen atoms in the ground state. The maximum value of $\lambda$ for which the hydrogen atom may be ionized is [take $hc = 1240 \text{ eV-nm}$]. (in $\text{ nm}$)