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

The light rays having photons of energy $4.2\,eV$ are falling on a metal surface having a work function of $2.2\,eV$. The stopping potential of the surface is $.........\,V$.

The stopping potential $V_0$ (in $volt$) as a function of frequency $(\nu)$ for a sodium emitter is shown in the figure. The work function of sodium,from the data plotted in the figure,will be: ................. $eV$
(Given: Planck's constant $(h) = 6.63 \times 10^{-34} \, Js$,electron charge $e = 1.6 \times 10^{-19} \, C$)

The work function of a metal is $4.2 \ eV$. Its threshold wavelength will be .......... $ \mathring A $.

The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of:

Light of wavelength $\lambda_{\text{ph}}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\phi$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. $A$ potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\lambda_e$,which of the following statement$(s)$ is(are) true?