Light of two different frequencies with photon energies of $1 \ eV$ and $2.5 \ eV$ are incident one after another on a metal surface having a work function of $0.5 \ eV$. The ratio of the maximum kinetic energy of the emitted photoelectrons will be:

$A$ photoelectric cell is connected to a source of variable potential difference. The photoelectric current is plotted against the applied potential difference. The graph with the broken line represents the current for a given frequency and intensity of the incident radiation. If the frequency is increased and the intensity is reduced, which curve now represents the situation?

The work function of sodium is $2.3 eV$. The threshold wavelength of sodium will be ........... $\mathring A $

In the photoelectric effect, the kinetic energy of the electron emitted from the metal surface depends upon:

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

Radiation $p$ with wavelength $6561 \; \mathring{A}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} \; T$. If the radius of the largest circular path followed by the electrons is $10 \; mm$,the work function of the metal is close to ............... $eV$.