**Hint:**

The energy of a photon is given by the relation:

$ E = \frac{hc}{\lambda} $

Where $ H $ is the Planck's constant $((6.626 \times 10^{-34}) J.s), ( c )$ is the speed of light $((3 \times 10^8) m/s), and ( \lambda )$ is the wavelength of the photon.

The kinetic energy (K.E) of the emitted photoelectron can be determined from the photoelectric equation:

$K.E = E - \text{Work function} $

The velocity $( v )$ of the photoelectron can be obtained using the relation:

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

where $( m )$ is the mass of the electron $((9.11 \times 10^{-31}) kg)$.

Step-by-Step Solution:

The energy of the photon:

$\begin{align*} E &= \frac{hc}{\lambda} \\ E &= \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} \\ E &= 4.9695 \times 10^{-19} \text{ J} \end{align*}$

Convert this energy to

$eV: \begin{align*} E &= \frac{4.9695 \times 10^{-19}}{1.6020 \times 10^{-19}} \\ E &\approx 3.10 \text{ eV} \end{align*}$

The kinetic energy of the emission:

$ K.E = E - \text{Work function} $ $\begin{align*} K.E. &= 3.10 - 2.13 \\ K.E. &\approx 0.97 \text{ eV} \end{align*}$

**The velocity of the photoelectron:**

**From the kinetic energy:**

$\begin{align*} K.E. &= \frac{1}{2} mv^2 \\ 0.97 \times 1.6020 \times 10^{-19} &= \frac{1}{2} \times 9.11 \times 10^{-31} \times v^2 \end{align*}$

$\text {Solving for} \ ( \nu): \begin{equation*} v \approx 5.85 \times 10^5 \text{ m/s} \end{equation*}$

**Answers:**

(i) The energy of the photon is approximately $3.10 \text{ eV}$

(ii) The kinetic energy of the emission is approximately $0.97 \text{ eV}$.

(iii) The velocity of the photoelectron is approximately $5.85 \times 10^5 \text{ m/s}$.

**Note:** The photoelectric effect describes the emission of photoelectrons from a material when it is exposed to electromagnetic radiation. The kinetic energy of these emitted electrons depends on the energy of the incident photons and the work function of the material. If the energy of the photon is less than the work function, no photoelectrons are emitted.