The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface,having work function $5.01 \, eV$,when ultraviolet light of $200 \, nm$ falls on it,must be ............... $V$.

When photons of energy $hf$ fall on an aluminium plate (of work function $E_0$),photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be:

$A$ photosensitive surface made of cesium on tungsten is irradiated by monochromatic radiation of wavelength $640.2 \ nm$ $(1 \ nm = 10^{-9} \ m)$ from a neon bulb. The measured stopping potential is $0.54 \ V$. If the source is replaced by another source and the same photocell is irradiated by a line of $427.2 \ nm$,what will be the new stopping potential in $V$?

According to Einstein's photoelectric equation,the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be:

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?