If a surface has a work function $4.0eV$, what is the maximum velocity of electrons liberated from the surface when it is irradiated with ultraviolet radiation of the wavelength $0.2\mu m$ ?
A) $4.4 \times {10^5}m{s^{ - 1}}$
B) $8.8 \times {10^7}m{s^{ - 1}}$
C) $8.8 \times {10^5}m{s^{ - 1}}$
D) $4.4 \times {10^7}m{s^{ - 1}}$

Hint: Under the right circumstances light can be used to push electrons, freeing them from the surface of a solid. This process is called the photoelectric effect. The velocity, thus, the kinetic energy and the energy of the liberated electron are dependent on each other. In order to find the value of the maximum velocity of the electron particle, just compare the two formulas which describe the value of the energy of the electron particle.

Complete step by step solution:
Here we need to find the value of the maximum velocity of the electron particle. First, define the terms given in the question:
Work function, $W = 4.0eV$
Wavelength of ultraviolet radiation, $\lambda = 0.2\mu m$
From Einstein’s photoelectric theory we know that, the energy of the electron,
$E = W + {(K.E)_{\max }}$…………………………………………………… (1)
Here, $W = 4.0eV = 4 \times 1.6 \times {10^{ - 19}}J$
${(K.E)_{\max }}$ is the maximum kinetic energy.
Let’s find out the value of the energy of electron,
$E = \dfrac{{hc}}{\lambda }$……………………………………………… (2)
Where, $h$ is the Planck's constant and which is given by,
$h = 6.6 \times {10^{ - 34}}{m^2}kg{s^{ - 1}}$
$c$ is the velocity of light, and is given by,
$c = 3 \times {10^8}m{s^{ - 1}}$
$\lambda $ is the wavelength of the radiation,
$\lambda = 0.2\mu m = 0.2 \times {10^{ - 6}}m$
We are applying these known values to the equation (2)
We get, $E = \dfrac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{0.2 \times {{10}^{ - 6}}}}$
$ \Rightarrow E = 3.3 \times 3 \times {10^{ - 34}} \times {10^{8 + 7}}$
$ \Rightarrow E = 9.9 \times {10^{ - 19}}J$
We are converting this value from $J$ into $ev$
$ \Rightarrow E = \dfrac{{9.9 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}ev$
$ \Rightarrow E = 6.1875eV$
Now we know the value of energy of the electron, $E = 6.1875eV$
Applying this value of energy of electron in the equation (2)
We get, $E = W + {(K.E)_{\max }}$
Applying the value of energy of the electron, $E = 6.1875eV$
And the value of work function, $W = 4.0eV$
We get, $E = 6.1875 = 4 \times 1.6 \times {10^{ - 19}} + {(K.E)_{\max }}$
$ \Rightarrow {(K.E)_{\max }} = 6.1875 - 4$
$ \Rightarrow {(K.E)_{\max }} = 2.1875eV$
$ \Rightarrow {(K.E)_{\max }} = 2.1875 \times 1.6 \times {10^{ - 19}}J$

Now we have the value of the maximum kinetic energy of the electron. We can calculate the value of the maximum velocity from the equation for the maximum kinetic energy.
It is known that the kinetic energy of a particle is given by the equation,
${(K.E)_{\max}}=\dfrac{1}{2}mv_{\max }^2$ ……………………………………….. (3)
Here, $m$ is the mass of the electron and is given by, $m = 9.1 \times {10^{ - 31}}kg$
Applying all the known values to the equation (3), we get,
$ \Rightarrow {(K.E)_{\max }} = 2.1875 \times 1.6 \times {10^{ - 19}}J = \dfrac{1}{2} \times 9.1 \times {10^{ - 31}}v_{\max }^2$
\[ \Rightarrow v_{\max }^2 = \dfrac{{2 \times 2.1875 \times 1.6 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}\]
\[ \Rightarrow v_{\max }^{} = \sqrt {\dfrac{{2 \times 2.1875 \times 1.6 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}} \]
\[ \Rightarrow v_{\max }^{} = 8.76 \times {10^5}m{s^{ - 1}}\]
\[ \Rightarrow v_{\max }^{} \approx 8.8 \times {10^5}m{s^{ - 1}}\]

The final answer is the Option (C): $8.8 \times {10^5}m{s^{ - 1}}$.

Note: Einstein's explanation of the photoelectric effect was very simple. He assumed that the kinetic energy of the ejected electron was equal to the energy of the incident photon minus the energy required to remove the electron from the material, which is called the work function.