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

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):

$A$ photon of energy $hv$ is absorbed by a free electron of a metal having work function $\phi < hv$.

In a photoelectric experiment,a graph of maximum kinetic energy $(KE_{\max})$ against the frequency of incident radiation $(\nu)$ is plotted. If $A$ and $B$ are the intercepts on the $X$ and $Y$ axis respectively,the Planck's constant is given by:

When light is incident on a surface,photoelectrons are emitted. For these photoelectrons,which of the following is true?

In a photoelectric effect experiment,the cathode metal is exposed to light of wavelength $600 \ nm$. The maximum kinetic energy of the ejected electron doubles when light of wavelength $400 \ nm$ is used. The work function of the cathode metal is approximately: [Use $h=6.63 \times 10^{-34} \ J-s, c=3 \times 10^8 \ m/s$] (in $eV$)