When the work function of a metal increases,the maximum kinetic energy of the emitted photoelectrons:

The electric field of a light wave is given as $\vec E = 10^{-3} \cos \left( \frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t \right) \hat x \, N/C$. This light falls on a metal plate with a work function of $2 \, eV$. The stopping potential of the photoelectrons is ................ $V$.

When radiation of wavelength $\lambda$ is incident on a metal,the stopping potential of photoelectrons is $4.8 \ V$. When radiation of wavelength $2\lambda$ is incident on the same metal,the stopping potential is $1.6 \ V$. What is the threshold wavelength of the metal?

When a light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are emitted. For another emitter,light of wavelength $600 \ nm$ is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

Light of wavelength $\lambda$ falls on a metal having work function $\frac{hc}{\lambda_0}$. Photoelectric effect will take place only if ($\lambda_0$ is the threshold wavelength).

The electric field associated with a light wave is given by $E = E_0 \sin [1.57 \times 10^7 (x - ct)]$,where $x$ is in meters and $t$ is in seconds. If this light is used to produce photoelectric emission from a metal surface with a work function of $1.9 \ eV$,what will be the stopping potential in $V$?