For a photosensitive material, the work function is $W_0$ and the stopping potential is $V$. What is the wavelength of the incident radiation? ($h$ = Planck's constant, $c$ = velocity of light, $e$ = electronic charge)

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:

An isolated metallic sphere is illuminated by light of $4\,eV$ photon energy. If the work function of the metal is $2\,eV$,then the minimum potential of the sphere so that no photoelectrons are ejected is ................ $V$.

The stopping potential for photoelectrons:

Light of frequency $4 \times 10^{14} \,Hz$ is incident on a metal surface of work function $2.14 \,eV$, resulting in photoemission of electrons. The maximum kinetic energy of the emitted electrons is $\left[h=6.63 \times 10^{-34} \,J-s\right]$ (in $\,eV$)

When a photosensitive metal surface is illuminated with radiation of wavelength $\lambda_1$,the stopping potential is $V_1$. If the same surface is illuminated with radiation of wavelength $3\lambda_1$,the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is: