The work function for a caesium atom is $1.9 \, eV$. Calculate $(a)$ the threshold wavelength and $(b)$ the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength of $500 \, nm$,calculate the kinetic energy and the velocity of the ejected photoelectron.

It is given that the work function $(W_{0})$ for a caesium atom is $1.9 \, eV$.
$(a)$ From the expression $W_{0} = \frac{hc}{\lambda_{0}}$,we get:
$\lambda_{0} = \frac{hc}{W_{0}}$
Where,$\lambda_{0} =$ threshold wavelength,$h =$ Planck's constant,$c =$ velocity of light.
Substituting the values:
$\lambda_{0} = \frac{(6.626 \times 10^{-34} \, Js)(3.0 \times 10^{8} \, ms^{-1})}{1.9 \times 1.602 \times 10^{-19} \, J} = 6.53 \times 10^{-7} \, m = 653 \, nm$.
$(b)$ From the expression $W_{0} = hv_{0}$,we get:
$v_{0} = \frac{W_{0}}{h} = \frac{1.9 \times 1.602 \times 10^{-19} \, J}{6.626 \times 10^{-34} \, Js} = 4.593 \times 10^{14} \, s^{-1}$.
$(c)$ For irradiation with $\lambda = 500 \, nm$:
Kinetic energy $(K.E.) = \frac{hc}{\lambda} - W_{0} = hc \left(\frac{1}{\lambda} - \frac{1}{\lambda_{0}}\right)$
$K.E. = (6.626 \times 10^{-34} \, Js)(3.0 \times 10^{8} \, ms^{-1}) \left(\frac{1}{500 \times 10^{-9} \, m} - \frac{1}{653 \times 10^{-9} \, m}\right) = 9.315 \times 10^{-20} \, J$.
Since $K.E. = \frac{1}{2} mv^{2}$,where $m = 9.109 \times 10^{-31} \, kg$:
$v = \sqrt{\frac{2 \times K.E.}{m}} = \sqrt{\frac{2 \times 9.315 \times 10^{-20} \, J}{9.109 \times 10^{-31} \, kg}} = 4.52 \times 10^{5} \, ms^{-1}$.