Light comprising three wavelengths $310 \ nm$,$455 \ nm$,and $620 \ nm$ is incident on a surface separating two media at an angle of $45^o$. The refractive index for the $455 \ nm$ light is $\sqrt{2}$. If a metal plate with a work function of $1.2 \ eV$ is placed in the second medium,what is the maximum kinetic energy of the electrons emitted from the metallic plate in $eV$?

When a photon of energy $4.25 \, eV$ strikes the surface of a metal $A$,the ejected photoelectrons have a maximum kinetic energy $T_A \, eV$ and de-Broglie wavelength ${\lambda _A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.70 \, eV$ is $T_B = (T_A - 1.50) \, eV$. If the de-Broglie wavelength of these photoelectrons is ${\lambda _B} = 2{\lambda _A}$,then:

Which of the following statements is not true about stopping potential $(V_0)$?

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $3V_0$. When the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

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$)

The work function of a metal is $1.6 \times 10^{-19} \, J$. When the metal surface is illuminated by light of wavelength $6400 \, \mathring{A}$,the maximum kinetic energy of the emitted photo-electrons will be. (Planck's constant $h = 6.4 \times 10^{-34} \, Js$)