The anode voltage of a photocell is kept fixed. The frequency of the light falling on the cathode is gradually increased. Then the correct graph which shows the variation of photo current $I$ with the frequency $\nu$ of incident light is

$A$ photocell stops emission if it is maintained at $2V$ negative potential. The energy of the most energetic photoelectron is:

Threshold frequency for a metal is $15 \times 10^{14} \, Hz$. The light of wavelength $6000 \, \text{Å}$ falls on the metal surface. Then photoelectrons [velocity of light in air, $c = 3 \times 10^8 \, m/s$]:

$A$ photosensitive material is at $9\,m$ to the left of the origin and the source of light is at $7\,m$ to the right of the origin along the $x$-axis. The photosensitive material and the source of light start from rest and move,respectively,with $8\widehat{i}\,m/s$ and $4\widehat{i}\,m/s$. The ratio of intensities at $t = 0$ to $t = 3\,s$ as received by the photosensitive material is:

In a photoelectric experiment,if the intensity of incident light is doubled and the frequency is kept slightly greater than threshold frequency,then the saturation photoelectric current

The curve between the work function of a metal $(\phi_o)$ and its temperature $(T)$ will be: