Five elements $A, B, C, D$ and $E$ have work functions $1.2 \, eV, 2.4 \, eV, 3.6 \, eV, 4.8 \, eV$ and $6 \, eV$ respectively. If light of wavelength $4000 \, Å$ is allowed to fall on these elements, then photoelectrons are emitted by:

$A$ photon of $5.5 \ eV$ energy falls on the surface of a metal,emitting photoelectrons with a maximum kinetic energy of $4.0 \ eV$. The stopping voltage required for these electrons is .......... $V$.

The electric field at the point associated with a light wave is given by $E = 200 [\sin(6 \times 10^{15} t) + \sin(9 \times 10^{15} t)] \, Vm^{-1}$. Given $h = 4.14 \times 10^{-15} \, eVs$. If this light falls on a metal surface having a work function of $2.50 \, eV$,the maximum kinetic energy of the photoelectrons will be ........... $eV$.

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency) is incident on a metal plate,the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$,the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ 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

Photoelectric effect can be explained by