When $UV$ light of wavelength $300 \, nm$ is incident on a metal surface having a work function of $2.13 \, eV$, electron emission takes place. The stopping potential is: (Given $hc = 1240 \, eV \cdot nm$) (in $ \, V$)

In a photoemission experiment,the maximum kinetic energies of photoelectrons from metals $P, Q$ and $R$ are $E_P, E_Q$ and $E_R$,respectively,and they are related by $E_P = 2E_Q = 2E_R$. In this experiment,the same source of monochromatic light is used for metals $P$ and $Q$,while a different source of monochromatic light is used for metal $R$. The work functions for metals $P, Q$ and $R$ are $4.0 \ eV$,$4.5 \ eV$ and $5.5 \ eV$,respectively. The energy of the incident photon used for metal $R$,in $eV$,is:

Radiation $p$ with wavelength $6561 \; \mathring{A}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} \; T$. If the radius of the largest circular path followed by the electrons is $10 \; mm$,the work function of the metal is close to ............... $eV$.

$A$ and $B$ are two metals with threshold frequencies $1.8 \times 10^{14} \ Hz$ and $2.2 \times 10^{14} \ Hz$. Two identical photons of energy $0.825 \ eV$ each are incident on them. Then photoelectrons are emitted by (Take $h = 6.6 \times 10^{-34} \ J \cdot s$)

Yellow light of wavelength $557 \ nm$ is incident on a cesium surface. When the cathode-anode voltage drops below $0.25 \ V$,no photoelectrons flow in the circuit. What is the threshold wavelength of the cesium surface in $nm$?

The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?