When a photon of light collides with a metal surface, the number of electrons (if any) coming out is:

$A$ 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).

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy,for the same wavelength of incident radiation?

$A$ beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10\%$ of the incident photons effectively produce photoelectrons,then find the current due to these electrons in $\mu A$. (Given: $hc = 1240\,eV\cdot nm$,$e = 1.6 \times 10^{-19}\,C$)

Photons of energy $7 \,eV$ are incident on two metals $A$ and $B$ with work functions $6 \,eV$ and $3 \,eV$,respectively. The minimum de-Broglie wavelengths of the emitted photoelectrons with maximum kinetic energies are $\lambda_A$ and $\lambda_B$ respectively,where $\lambda_A / \lambda_B$ is nearly

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.