In the photoelectric emission process from a metal with a work function of $1.8\, eV$,the kinetic energy of the most energetic electrons is $0.5\, eV$. The corresponding stopping potential is ......... $V$.

Photoelectric effect can be explained by

When a certain metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the surface is:

When a photosensitive surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

Light of frequency $4\nu_0$ is incident on a metal surface with a threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is: