In an experiment on the photoelectric effect,the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} \; V \cdot s$. Calculate the value of Planck's constant.

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

Why maximum kinetic energy of a photoelectron cannot be negative?

The beam of light has three wavelengths $4144 \,\mathring A$,$4972 \,\mathring A$,and $6216 \,\mathring A$ with a total intensity of $3.6 \times 10^{-5} \, W/m^2$ equally distributed amongst the three wavelengths. The beam falls normally on an area of $1 \, cm^2$ of a clean metallic surface with a work function of $2.3 \, eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2 \, s$.

In the experiment of the photoelectric effect, if the frequency of incident light is $v_1$, the maximum kinetic energy $(K.E.)$ of photoelectrons is $K_0$. If the frequency of light is $v_2$, the maximum $K.E.$ of photoelectrons is $2K_0$. Which of the following relations is correct?