From the principle of conservation of energy, the energy of an incident photon (E) is equal
to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,
E = ${\mathrm{W}}_0$ + K.E
⇒ ${\mathrm{W}}_0$ = E – K.E
The energy of the incident photon (E) = $= \frac{\mathrm{hc}}{\mathrm{\lambda }}$
Where, c = velocity of radiation
h = Planck’s constant
λ = wavelength of radiation
Substituting the values in the given expression of E:

$\mathrm{E} = \frac{\left(6.626 \mathrm{x} {10}^{-34} \mathrm{Js}\right)\left(3.0 \mathrm{x} {10}^8 \mathrm{m}/\mathrm{s}\right)}{256.7 \mathrm{x} {10}^{-9} \mathrm{m}}$
$= 7.744 \mathrm{x} {10}^{-19} \mathrm{J}$
$= \frac{7.744 \mathrm{x} {10}^{-19}}{1.602 \mathrm{x} {10}^{-19}} \mathrm{eV}$

E = 4.83 eV
The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron.
Hence,
K.E = 0.35 V
K.E = 0.35 eV
The work function, W0 = E – K.E
= 4.83 eV – 0.35 eV
= 4.48 eV

$\frac{5{\mathrm{\lambda }}_0 - 2000}{4{\mathrm{\lambda }}_0 - 2000} = {\left(\frac{5.35}{2.55}\right)}^2 = \frac{28.6225}{6.5025}$
$\frac{5{\mathrm{\lambda }}_0 - 2000}{4{\mathrm{\lambda }}_0 - 2000} = 4.40177$
$17.6070{\mathrm{\lambda }}_0 - 5{\mathrm{\lambda }}_0 = 8803.537 - 2000$
${\mathrm{\lambda }}_0 = \frac{6805.537}{12.607}$
${\mathrm{\lambda }}_0 = 539.8 \mathrm{nm}$
${\mathrm{\lambda }}_0 = 540 \mathrm{nm}$