When light of frequency $v$ is incident on a metal,photoelectrons are emitted. This is because...

When radiations of wavelength $\lambda$ are incident on a metallic surface,the stopping potential required is $4.8 \ V$. If the same surface is illuminated with radiations of double the wavelength,the required stopping potential becomes $1.6 \ V$. The value of the threshold wavelength for the surface is:

Ultraviolet light of wavelength $2271 \,\mathring{A}$ from a $100 \; W$ mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is $-1.3 \; V$, estimate the work function of the metal. How would the photo-cell respond to a high-intensity $(10^{5} \; W \; m^{-2})$ red light of wavelength $6328 \,\mathring{A}$ produced by a $He-Ne$ laser?

When a metal plate is illuminated with light of wavelengths $400 \ nm$ and $250 \ nm$,the maximum velocities of the emitted photoelectrons are $v$ and $2v$,respectively. The work function of the metal is ($h$ = Planck's constant; $c$ = speed of light in vacuum):

$A$ metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value,then the maximum $KE$ of the emitted photoelectrons would be

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is: