Light of frequency $v$ falls on a material of threshold frequency $v_0$. Maximum kinetic energy of emitted electron is proportional to

$A$ metal surface having work function '$W_{0}$' emits photoelectrons when photons of energy '$E$' are incident on it. The electron enters a uniform magnetic field '$B$' in a perpendicular direction and moves in a circular path of radius '$r$'. Then '$r$' is equal to (where '$m$' and '$e$' are the mass and charge of the electron,respectively).

$A$ photoelectric cell is connected to a source of variable potential difference and the resulting photoelectric current $(\mu A)$ is plotted against the applied potential difference $(V)$. The graph with the broken line represents one situation for a given frequency and intensity of the incident radiation. If the frequency is increased and intensity is reduced,which of the following graphs of unbroken line represents the new situation?

The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

The graph between the maximum speed $v_{\text{max}}$ of a photoelectron and frequency $(\nu)$ of the incident radiation,in the photoelectric effect,is correctly represented by:

When photons of energy $hv$ fall on a metal plate of work function $W_0$,photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of the radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be