The maximum velocity of the photoelectron emitted by the metal surface is $v$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is $6250 \ \text{Å}$, then the work function of the metal is (Planck's constant $= 6.6 \times 10^{-34} \ \text{Js}$) (in $\text{eV}$)

The threshold wavelength for photoelectric emission from a material is $5200 \, \mathring{A}$. Photo-electrons will be emitted when this material is illuminated with monochromatic radiation from a

When a certain metal surface is illuminated with light of frequency $v$,the stopping potential for the photoelectric current is $V_s$. When the same surface is illuminated by a light of frequency $v/2$,the stopping potential is $V_s/4$. The threshold frequency of photoelectric emission is

Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 \, nm$. If the stopping potential for the ejected electrons is $2.8 \, eV$, then the work function of the metal is (Take $hc = 1240 \, eV \cdot nm$). (in $ \, eV$)

Ultraviolet light of wavelength $200 \ nm$ is incident on a freshly polished surface of iron. The work function of the surface is $4.71 \ eV$. What will be the stopping potential in $eV$? $(h = 6.626 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J, c = 3 \times 10^8 \ m/s)$