Photoelectric effect can be explained by

Light of incident frequency $2$ times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{3}\right)^{\text{rd}}$ and intensity is doubled,then the photoelectric current will

Two photons of energy $2.5 eV$ and $3.5 eV$ fall on a metal surface of work function $1.5 eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is

Find the number of electrons emitted per second by a $24 \, W$ source of monochromatic light of wavelength $6600 \, \mathring{A}$, assuming $3 \%$ efficiency for the photoelectric effect (take $h = 6.6 \times 10^{-34} \, J \cdot s$ and $c = 3 \times 10^8 \, m/s$).

The stopping potential for a photoelectric emission process is $10 \ V$. The maximum kinetic energy of the electrons ejected in the process is [Charge on electron $e = 1.6 \times 10^{-19} \ C$]