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$)

$A$ 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).

$A$ stream of photons having energy $3 \,eV$ each impinges on a potassium surface. The work function of potassium is $2.3 \,eV$. The emerging photo-electrons are slowed down by a copper plate placed $5 \,mm$ away. If the potential difference between the two metal plates is $1 \,V$,the maximum distance the electrons can move away from the potassium surface before being turned back is .......... $mm$.

When photons of energy $1 \ eV$ and $2.5 \ eV$ are incident on a metal surface with a work function of $0.5 \ eV$,what is the ratio of the maximum kinetic energies of the emitted photoelectrons?

The work function of a certain metal is $3.31 \times 10^{-19} \,J$. The maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \text{ Å}$ is (Given: $h = 6.62 \times 10^{-34} \,J \cdot s$,$c = 3 \times 10^8 \,m/s$,$e = 1.6 \times 10^{-19} \,C$) (in $\text{ eV}$)

$A$ photon of energy $hv$ is absorbed by a free electron of a metal having work function $\phi < hv$.