Explanation

(a) From Planck’s quantum theory, the maximum kinetic energy (E$_{k}$) of photoelectron emitted from the metal is
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E}_{k} = hv - W$
Where W is work function of metal and$\ $hv is the energy of the photon absorbed by the metal.
Given, $hv = 2\ eV,\ \ W = 1.4\ eV$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{k} = 2 - 1.4 = 0.6$eV
Hence, stopping potential is , $V_{s} = \frac{E}{e} = \frac{0.6\ eV}{e} = 0.6\ V.\ $