The difference between threshold wavelengths for two metal surfaces $A$ and $B$ having work functions $\phi_A = 9 \, eV$ and $\phi_B = 4.5 \, eV$ in $nm$ is: (Given,$hc = 1242 \, eV \, nm$)

The work function of caesium metal is $2.14 \; eV$. When light of frequency $6 \times 10^{14} \; Hz$ is incident on the metal surface,photoemission of electrons occurs. What is the
$(a)$ maximum kinetic energy of the emitted electrons,
$(b)$ stopping potential,and
$(c)$ maximum speed of the emitted photoelectrons?

The threshold frequency of a metal is $f_0$. When light of frequency $2f_0$ is incident on the metal plate,the maximum velocity of the photoelectrons is $v_1$. When the frequency of the incident radiation is increased to $5f_0$,the maximum velocity of the photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is:

The surface of a metal is illuminated alternately with photons of energies $E_{1} = 4 \ eV$ and $E_{2} = 2.5 \ eV$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is $2$. The work function of the metal in $eV$ is:

The work function of a metal is $2 \ eV$. If a radiation of wavelength $3000 \ \text{Å}$ is incident on it,the maximum kinetic energy of the emitted photoelectrons is (Planck's constant $h=6.6 \times 10^{-34} \ \text{Js}$; velocity of light $c=3 \times 10^8 \ \text{m/s}$; $1 \ \text{eV}=1.6 \times 10^{-19} \ \text{J}$).

Energy of a photoelectron depends on which factors?