When radiation of frequency $6 \times 10^{14} \ Hz$ is incident on a metal surface with a work function of $2 \ eV$,the maximum kinetic energy of the emitted photoelectrons is ........ $eV$ (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$,$1 \ eV = 1.6 \times 10^{-19} \ J$). (in $.49$)

In the photoelectric effect,a graph is plotted between the maximum kinetic energy of electrons and the frequency of incident photons. The slope of this curve will be .......

The threshold frequency of a metal is $F_0$. When light of frequency $2F_0$ is incident on the metal plate,the maximum velocity of the photoelectrons is $V_1$. When the frequency of incident radiation is increased to $5F_0$,the maximum velocity of the photoelectrons emitted is $V_2$. The ratio of $V_1$ to $V_2$ is:

Light of frequency $v$ is incident on a certain photoelectric material with a threshold frequency $v_0$. What is the work function for the material?

Which one of the following four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident light $(I)$ is correct?

If the stopping potential of photoelectrons emitted by a source placed at a distance of $0.2 \ m$ from a photocell is $0.6 \ V$,then the stopping potential of photoelectrons emitted by the same source placed at a distance of $0.6 \ m$ from the photocell will be .......... $V$.