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).

For a certain metal,when monochromatic light of wavelength $\lambda$ is incident,the stopping potential for photoelectrons is $3V_0$. When the same metal is illuminated by light of wavelength $2\lambda$,then the stopping potential becomes $V_0$. The threshold wavelength for photoelectric emission for the given metal is $\alpha\lambda$. The value of $\alpha$ is . . . . . . .

When light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are just liberated. 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

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

According to Einstein's photoelectric equation,the graph of the kinetic energy of photoelectrons emitted from a metal versus the frequency of incident radiation is a straight line whose slope:

Statement $-1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X$-rays, both $V_0$ and $K_{max}$ increase.
Statement $-2$: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.