When photons are incident on a photosensitive material of work function $1.5 \text{ eV}$,the maximum velocity of the emitted photoelectrons is $8 \times 10^5 \text{ m/s}$. The stopping potential of the photoelectrons is (Mass of the electron $= 9 \times 10^{-31} \text{ kg}$ and charge of the electron $= 1.6 \times 10^{-19} \text{ C}$) (in $V$)

When a photon of energy $4.0 \; eV$ strikes the surface of a metal $A$,the ejected photoelectrons have 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 $4.50 \; eV$ is $T_{B} = (T_{A} - 1.5) \; eV$. If the de-Broglie wavelength of these photoelectrons is $\lambda_{B} = 2 \lambda_{A}$,then the work function of metal $B$ is ............. $eV$.

For a metal with a work function of $6.6 \text{eV}$, which of the following wavelengths of incident radiation does not cause the photoelectric effect (in $\text{nm}$)? (Take Planck's constant $h = 6.6 \times 10^{-34} \text{J s}$ and speed of light $c = 3 \times 10^8 \text{m/s}$)

When a light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are emitted. For another emitter,light of wavelength $600 \ nm$ is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

Light of energy $E$ falls normally on a metal of work function $\frac{E}{3}$. The kinetic energies $K$ of the photoelectrons are

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential of ejected photoelectrons is $4.8 \, V$. If the same surface is illuminated by radiation of double the previous wavelength,then the stopping potential becomes $1.6 \, V$. The threshold wavelength of the metal is $... \lambda$.