In the graph given below,if the slope is $4.12 \times 10^{-15} \, V-s$,then the value of $'h'$ should be:

When a light of wavelength $4900 Å$ falls on a photosensitive metal, a negative $2 \,V$ potential is required to stop the emitted electrons. Then, the work-function of the material is nearly (given charge on electron $= 1.602 \times 10^{-19} C$ and Planck's constant $= 6.625 \times 10^{-34} Js$) (in $eV$)

The maximum velocity of the photoelectron emitted by the metal surface is $v$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volt is

On a photosensitive material,when the frequency of incident radiation is increased by $20 \%$,the maximum kinetic energy of emitted photoelectrons increases from $0.4 \ eV$ to $0.7 \ eV$. The work function of the material is (in $eV$)

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:

When monochromatic light is incident on a photosensitive surface,the number of photoelectrons emitted per second from the surface is $n$ and the maximum kinetic energy is $K_{max}$. If the intensity of the incident light is doubled,then...