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 $c = 3 \times 10^8 \text{m/s}$)

Two light waves of wavelengths $600 \,nm$ and $200 \,nm$ are incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is $\frac{1}{3}$ of the maximum velocity of the photoelectrons produced due to the other wavelength. The work function of the metal is:

The stopping potential for photoelectrons:

In a photoelectric experiment,the stopping potential was measured to be $V_1$ and $V_2$ volts with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The value of $V_2$ is: [where $\phi=$ work function,$e=$ electronic charge]

When photons of energy $h \nu$ fall on a photosensitive surface of work function $E_0$,photoelectrons of maximum kinetic energy $k$ are emitted. If the frequency of radiation is doubled,the maximum kinetic energy will be equal to ($h=$ Planck's constant).

When radiation of wavelength $\lambda$ is used to illuminate a metallic surface,the stopping potential is $V.$ When the same surface is illuminated with radiation of wavelength $3 \lambda,$ the stopping potential is $\frac{V}{4}.$ If the threshold wavelength for the metallic surface is $n \lambda,$ then the value of $n$ will be......