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

The curves $(1)$,$(2)$,$(3)$,and $(4)$ show the variation between the applied potential difference $(V)$ and the photoelectric current $(i)$ at two different intensities of light $(I_1 > I_2)$. In which figure is the correct variation shown?

Light of intensity $10^{-5} \; W m^{-2}$ falls on a sodium photo-cell of surface area $2 \; cm^2$. Assuming that the top $5$ layers of sodium absorb the incident energy,estimate the time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about $2 \; eV$. What is the implication of your answer?

The work function of a metal is

In the photoelectric effect,if a radiation of weak intensity is used instead of a radiation of strong intensity,but of the same suitable frequency,then:

The cathode of a photoelectric cell is changed such that the work function changes from $W_1$ to $W_2$ $(W_2 > W_1)$. If the current before and after change are $I_1$ and $I_2$,all other conditions remaining unchanged,then (assuming $h\nu > W_2$):