When radiations of wavelength $\lambda$ are incident on a metallic surface,the stopping potential required is $4.8 \ V$. If the same surface is illuminated with radiations of double the wavelength,the required stopping potential becomes $1.6 \ V$. The value of the threshold wavelength for the surface is:

If the work function of a metal is $\phi$ and the frequency of the incident light is $\nu$,there is no emission of photoelectrons if:

In a photoelectric experiment,light of energy $2.48 eV$ irradiates a photosensitive material. The stopping potential was measured to be $0.5 V$. The work function of the photosensitive material is: (in $eV$)

From the figure describing the photoelectric effect,we may infer correctly that:

When a photosensitive material is illuminated by photons of energy $3.1 \ eV$,the stopping potential of the photoelectrons is $1.7 \ V$. When the same photosensitive material is illuminated by photons of energy $2.5 \ eV$,the stopping potential of the photoelectrons is: (in $V$)

The wavelength of light in the visible region is about $390\; nm$ for violet colour,about $550\; nm$ (average wavelength) for yellow-green colour and about $760\; nm$ for red colour.
$(a)$ What are the energies of photons in $(eV)$ at the $(i)$ violet end,$(ii)$ average wavelength (yellow-green colour),and $(iii)$ red end of the visible spectrum? (Take $h=6.63 \times 10^{-34} \;J s$ and $1 \;eV = 1.6 \times 10^{-19} \;J$)
$(b)$ From which of the photosensitive materials with work functions listed in the table,and using the results of $(i), (ii)$ and $(iii)$ of $(a)$,can you build a photoelectric device that operates with visible light?

Metal	Work function $\phi_{0} (eV)$	Metal	Work function $\phi_{0} (eV)$
$Cs$	$2.14$	$Al$	$4.28$
$K$	$2.30$	$Hg$	$4.49$
$Na$	$2.75$	$Cu$	$4.65$
$Ca$	$3.20$	$Ag$	$4.70$
$Mo$	$4.17$	$N$	$5.15$
$Pb$	$4.25$	$Pt$	$5.65$