Answer
i) No photo electrons will be ejected.
ii) $K.E = 3.196\times 10^{-19}J, v = 837km/s$
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 $W_0 = 2.14eV = 3.43\times 10^{-19}J$
$i) \space \lambda = 700nm$
$\frac{hc}{\lambda} = \frac{6.626\times 10^{-34} \times 3\times 10^8}{700\times 10^{-9}} = 2.84 \times 10^{-19}J$
Since the energy of the incident photon is less than the Work function.
Hence, No photoelectrons will be ejected.
$ii) \space \lambda = 300nm$
$\frac{hc}{\lambda} = \frac{6.626\times 10^{-34} \times 3\times 10^8}{300\times 10^{-9}} = 6.626 \times 10^{-19}J$
Since the energy of the incident photon is greater than the Workfunction.
Hence, photoelectrons will be ejected.
$$K.E_{electron} = \frac{hc}{\lambda} - W_0$$
$$K.E_{electron} = 6.626\times 10^{-19} - 3.43\times 10^{-19} = 3.196\times 10^{-19}J$$
Also,
$v = \sqrt{\frac{2K.E}{m}} = \sqrt{\frac{2\times 3.196\times 10^{-19}}{9.11\times 10^{-31}}} = 837km/s$
