Light of frequency $4v_0$ is incident on a metal surface with a threshold frequency $v_0$. The maximum kinetic energy of the emitted photoelectrons is:

The work function of a metallic surface is $5.01 \, eV$. The photo-electrons are emitted when light of wavelength $2000 \, \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is ............... $volt$ $[h = 4.14 \times 10^{-15} \, eV \cdot s]$

For the photoelectric effect,the maximum kinetic energy $(E_{k})$ of the photoelectrons is plotted against the frequency $(\nu)$ of the incident photons as shown in the figure. The slope of the graph gives:

Photoelectric emission occurs only when the incident light has more than a certain minimum

Photons of wavelength $\lambda$ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius $R$ by a perpendicular magnetic field having a magnitude $B$. The work function of the metal is (where,symbols have their usual meanings)

The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of: