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

The work functions for tungsten and sodium are $4.5 \text{ eV}$ and $2.3 \text{ eV}$ respectively. If the threshold wavelength $\lambda_0$ for sodium is $5460 \text{ \AA}$, the value of $\lambda_0$ for tungsten is ............ $\text{ \AA}$.

For the photoelectric effect,the maximum kinetic energy $(E_{k})$ of the photoelectrons is plotted against the frequency $(\nu)$ of the incident photons as shown in the figure. The slope of the graph gives:

$A$ metal surface is illuminated by light of two different wavelengths $248 \ nm$ and $310 \ nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$,respectively. If the ratio $u_1: u_2 = 2: 1$ and $hc = 1240 \ eV \ nm$,the work function of the metal is nearly: (in $eV$)

When a point source of light is at a distance of $1 \ m$ from a photocell,the cut-off voltage is found to be $V$. If the same source is placed at $2 \ m$ distance from the photocell,the cut-off voltage will be

The work function of a metal is $1.6 \times 10^{-19} \, J$. When the metal surface is illuminated by light of wavelength $6400 \, \mathring{A}$,the maximum kinetic energy of the emitted photo-electrons will be. (Planck's constant $h = 6.4 \times 10^{-34} \, Js$)