The work function of caesium metal is $2.14 \; eV$. When light of frequency $6 \times 10^{14} \; Hz$ is incident on the metal surface,photoemission of electrons occurs. What is the
$(a)$ maximum kinetic energy of the emitted electrons,
$(b)$ stopping potential,and
$(c)$ maximum speed of the emitted photoelectrons?

(N/A) Given: Work function $\phi_{0} = 2.14 \; eV = 2.14 \times 1.6 \times 10^{-19} \; J = 3.424 \times 10^{-19} \; J$. Frequency $\nu = 6 \times 10^{14} \; Hz$. Planck's constant $h = 6.63 \times 10^{-34} \; J \cdot s$. Mass of electron $m = 9.11 \times 10^{-31} \; kg$.
$(a)$ Using Einstein's photoelectric equation,$K_{\max} = h\nu - \phi_{0}$.
$h\nu = (6.63 \times 10^{-34}) \times (6 \times 10^{14}) = 3.978 \times 10^{-19} \; J$.
$K_{\max} = 3.978 \times 10^{-19} - 3.424 \times 10^{-19} = 0.554 \times 10^{-19} \; J$.
In $eV$,$K_{\max} = (0.554 \times 10^{-19}) / (1.6 \times 10^{-19}) \approx 0.346 \; eV$.
$(b)$ Stopping potential $V_{0}$ is given by $eV_{0} = K_{\max}$.
$V_{0} = (0.554 \times 10^{-19} \; J) / (1.6 \times 10^{-19} \; C) = 0.346 \; V$.
$(c)$ Maximum speed $v_{\max}$ is given by $K_{\max} = \frac{1}{2}mv_{\max}^{2}$.
$v_{\max} = \sqrt{\frac{2K_{\max}}{m}} = \sqrt{\frac{2 \times 0.554 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{0.1216 \times 10^{12}} \approx 3.49 \times 10^{5} \; m/s = 349 \; km/s$.