Answer
i) No photoelectrons will be ejected.
ii) $K.E = 6.84\times 10^{-19}J, v = 1.23Mm/s$
Work Step by Step
Photoelectrons are ejected if the incident energy of photon is greater than the work function($W_0$) of the material &
According to Einstein's Photoelectric equation, Kinetic energy of ejected electron is given be
$$K.E_{electron} = \frac{hc}{\lambda} - W_0$$
Given $W_0 = 2.09eV = 3.35\times 10^{-19}J$
$i) \space \lambda = 650nm$
$\frac{hc}{\lambda} = \frac{6.626\times 10^{-34} \times 3\times 10^8}{650\times 10^{-9}} = 3.06 \times 10^{-19}J$
Since the energy of the incident photon is less than the Work function.
Hence, No photoelectrons will be ejected.
$ii) \space \lambda = 195nm$
$\frac{hc}{\lambda} = \frac{6.626\times 10^{-34} \times 3\times 10^8}{195\times 10^{-9}} = 10.19 \times 10^{-19}J$
Since the energy of the incident photon is greater than the Workfunction.
Hence, photoelectrons will be ejected.
$$K.E_{electron} = \frac{hc}{\lambda} - W_0$$
$$K.E_{electron} = 10.19\times 10^{-19} - 3.35\times 10^{-19} = 6.84\times 10^{-19}J$$
Also,
$v = \sqrt{\frac{2K.E}{m}} = \sqrt{\frac{2\times 6.84\times 10^{-19}}{9.11\times 10^{-31}}} = 1225km/s = 1.23Mms^{-1}$
