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Formula used: $n = \dfrac{{\text{P}}}{E}$

\[h\nu = {\text{ }}{\phi _0} + {K_{\max }}\]

$E = h\nu $

$c = \nu \lambda $

Where,P is the power, h is the Planck’s constant $h = 6.625 \times {10^{ - 34}}Js$, E is the energy of the photon, c is the speed of light, $\nu $is the frequency of an incident light, \[{\phi _0}\]work function.

Let us find the value for the frequency of the electron.

Given, the wavelength of the ultra violet light, $\lambda = 2480\mathop A\limits^0 $, power $P = 40W$

Radius =$2m$

Work function$\phi = 3.68eV$

The energy of each photon in the ultraviolet source,$E = h\nu $

In this question value of frequency, $\nu$ is not given.

Therefore, we can write,

$c = \nu \lambda $

$\therefore \nu = \dfrac{c}{\lambda }$

We can compare the above equation with the Einstein and Plank relation

Consider $E = h\nu $

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

$E = \dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2480 \times {{10}^{ - 10}}}}$

$E = 8.008 \times {10^{ - 19}}{\text{J}}$

\[E = \dfrac{{8.008 \times {{10}^{ - 19}}J}}{{1.6 \times {{10}^{ - 19}}}}\]

\[ = 5.005eV\]

Therefore, the value of the energy of the photon is \[ = 5.005eV\]

Let us now consider the value of the number of photons emitted from the surface per second

$n = \dfrac{P}{E}$

$n = \dfrac{{40}}{{8.008 \times {{10}^{ - 19}}}}$

$n = 5 \times {10^{19}}photons{\text{ }}per{\text{ }}\text{\second}$

We can consider these photons is spread in all the directions over the magnesium surface area $4\pi {r^2}$.

and the number of electrons incident on unit area of Mg surface per second, $N = \dfrac{n}{{{\text{surface area}}}}$

$N = \dfrac{{5 \times {{10}^{19}}}}{{4\pi {r^2}}}$

$N = \dfrac{{5 \times {{10}^{19}}}}{{4 \times 3.14 \times \left( {{2^2}} \right)}}$

$ = 9.95 \times {10^{13}}per{\text{ }}\sec ond$

From Einstein’s photoelectric equation,

When the light is incident on metal, the photons having energy collide with electrons with the surface of the metal. During these collisions, the energy of the photon is completely transferred to the electron. If this energy is sufficient, the electrons are ejected out of the metal instantaneously. the minimum energy needed for the electron to come out of the metal surface is called work function. If the energy of the incident photon exceeds the work function, the electrons are emitted with maximum kinetic energy.

\[h\nu = {\text{ }}{\phi _0} + {K_{\max }}\]

\[{K_{\max }} = h\nu - {\phi _0}\]

${K_{\max }} = E - {\phi _0}$

${K_{\max }} = 5eV - 3.68eV$

${\text{ = }}1.32eV$

Maximum wavelength for photoelectric emission that is Threshold wavelength $\left( {{{{\lambda }}_{{0}}}} \right)$.

Threshold wavelength is defined as the certain maximum wavelength of the incident radiation required for the emission of photoelectrons. This maximum wavelength is called the threshold wavelength.

Work function

\[{\phi _0} = h{v_0}\]

\[{\phi _0} = \dfrac{{hc}}{{{\lambda _0}}}\]

\[{\lambda _0} = \dfrac{{hc}}{{{\phi _0}}}\]

\[{\phi _0} = \dfrac{{6.625 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3.68 \times 1.6 \times {{10}^{ - 19}}}}\] \[ = 3375\mathop {\text{A}}\limits^0 \]

\[\dfrac{{F.{Q_1}{Q_2}}}{{{d^2}}}\]

\[where,F = 9 \times {10^{19}}N\dfrac{{{m^2}}}{{{c^2}}}\]

\[{Q_1} = 1.6 \times {10^{ - 16}}c\]

\[{Q_2} = 1.6 \times {10^{ - 16}}c\]

\[d = 0.53 \times {10^{ - 10}}m\]

\[F = \dfrac{{9 \times {{10}^{19}}N\dfrac{{{m^2}}}{{{c^2}}} \times 1.6 \times {{10}^{ - 16}}c \times 1.6 \times {{10}^{ - 16}}c}}{{{{(0.53 \times {{10}^{ - 10}}m)}^2}}}\]

\[F = \dfrac{{23.04N}}{{0.2809}}\] \[F = 8.2 \times {10^{ - 8}}N\]

When the electromagnetic radiation hits a material, the electrons are emitted. The process is called the Photoelectric effect and the electrons thus emitted are called the Photo electrons. According to the wave theory light should have a suitable frequency to emit the electrons.

Note:

This particular problem involves many values. After analyzing each value given substitute those values in the formulae. The energy of the electron is the key formula from which the answer can be derived.