The work functions of metals $A$ and $B$ are in the ratio $1 : 2$. If light of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energy of photoelectrons emitted will be ($f$ and $2f$ are both frequencies greater than the threshold frequency of metals $A$ and $B$).

An electron in a hydrogen atom jumps from the second Bohr orbit to the ground state,and the energy difference of the two states is radiated in the form of photons. These photons are then allowed to fall on a metal surface having a work function equal to $4.2 \ eV$. Calculate the stopping potential. [Energy of electron in $n^{\text{th}}$ orbit $= -\frac{13.6}{n^2} \ eV$] (in $V$)

$A$ photosensitive metallic surface has a work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface,the electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. When the photon energy is increased to $5h\nu_0$,the maximum velocity of the photoelectrons will be:

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.

An electron in a Hydrogen atom jumps from the second Bohr orbit to the ground state, and the difference between the energies of the two states is radiated in the form of a photon. This photon strikes a material. If the work function of the material is $4.2 \ eV$, then the stopping potential is (Energy of electron in $n$-th orbit $= -\frac{13.6}{n^2} \ eV$). (in $V$)

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is