If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photoelectrons would be in joule:

Light of wavelength $4000 \,\mathring A$ falls on a photosensitive metal and a negative $2 \,V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 \times 10^{-34} \,Js, \,e = 1.6 \times 10^{-19} \,C, \,c = 3 \times 10^8 \,m/s)$.

In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

The frequency and the intensity of a beam of light falling on the surface of a photoelectric material are both increased by a factor of two. This will:

The photoelectric threshold wavelength for silver is $\lambda_{0}$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda$ (where $\lambda < \lambda_{0}$) will be:

When a certain metal surface is illuminated with light of frequency $v$,the stopping potential for photoelectric current is $V_0$. When the same surface is illuminated by light of frequency $\frac{v}{2}$,the stopping potential is $\frac{V_0}{4}$,the threshold frequency of photoelectric emission is