The work function of a metallic surface is $5.01 \, eV$. The photo-electrons are emitted when light of wavelength $2000 \, \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is ............... $volt$ $[h = 4.14 \times 10^{-15} \, eV \cdot s]$

When a point source of light is at a distance of $50 \, cm$ from a photoelectric cell,the cut-off voltage is found to be $V_0$. If the same source is placed at a distance of $1 \, m$ from the cell,then the cut-off voltage will be

Sodium and copper have work functions $2.3 \ eV$ and $4.5 \ eV$ respectively. Then the ratio of their threshold wavelengths is nearest to

$A$ metal surface is illuminated by light of a given intensity and frequency, causing photoelectric emission. If the intensity of the incident light is reduced to $1/4$ of its original value, what will be the kinetic energy of the emitted photoelectrons?

When radiation is incident on a photoelectron emitter,the stopping potential is found to be $9 \ V$. If $e/m$ for the electron is $1.8 \times 10^{11} \ C \ kg^{-1}$,the maximum velocity of the ejected electrons is:

Electrons ejected from the surface of a metal,when light of a certain frequency is incident on it,are stopped fully by a retarding potential of $3 \ V$. The photoelectric effect on this metallic surface begins at a frequency of $6 \times 10^{14} \ s^{-1}$. The frequency of the incident light in $s^{-1}$ is: [Planck's constant $= 6.4 \times 10^{-34} \ J \cdot s$,charge on the electron $= 1.6 \times 10^{-19} \ C$]