The work function of caesium metal is 2.14 eV.

Work function of caesium metal, $\phi_{0}=2.14 \mathrm{eV}$

Frequency of light, $v=6.0 \times 10^{14} \mathrm{~Hz}$

(a)The maximum kinetic energy is given by the photoelectric effect as:

$K=h v-\phi_{0}$

Where,

h = Planck’s constant = 6.626 × 10−34 Js

$\therefore K=\frac{6.626 \times 10^{34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}-2.14$

$=2.485-2.140=0.345 \mathrm{eV}$

Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.

(b)For stopping potential $V_{0}$, we can write the equation for kinetic energy as:

$K=e V_{0}$

$\therefore V_{0}=\frac{K}{e}$

$=\frac{0.345 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=0.345 \mathrm{~V}$

Hence, the stopping potential of the material is 0.345 V.

(c)Maximum speed of the emitted photoelectrons = v

Hence, the relation for kinetic energy can be written as:

$K=\frac{1}{2} m v^{2}$

Where,

m = Mass of an electron = 9.1 × 10−31 kg

$v^{2}=\frac{2 K}{m}$

$=\frac{2 \times 0.345 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}=0.1104 \times 10^{12}$

$\therefore v=3.323 \times 10^{5} \mathrm{~m} / \mathrm{s}=332.3 \mathrm{~km} / \mathrm{s}$

Hence, the maximum speed of the emitted photoelectrons is
332.3 km/s.