Anode voltage is at ${{ + 3V}}$ . Incident radiation has frequency ${{1}}{{.4 \times 1}}{{{0}}^{{{15}}}}{Hz}$ and work function of the photo cathode is ${{2}}{{.8eV}}{{.}}$ find the minimum and maximum KE of photon electron in ${{eV}}{{.}}$
A) $3, 6$
B) $0, 3$
C) $0, 6$
D) $2.8,5.8$

Hint: In order to solve this question, we will use the formula ${{E = qV}}$ for calculating energy and remember that it is accelerating by anode voltage that is ${{3V}}$ energy is added to energy.
This problem is an example of Einstein’s photoelectric emission.

Formula used:
 ${{E = qV}}$
Where ${{E = }}$ energy due to accelerating voltage
${{q}}$ is equal to charge
${{\& }}{{V}}$ is voltage.
${{h\upsilon }}{{ = }}{{f + K}}{{{E}}_{{{max}}}}$
Where
${{h}}$ is planck's constant
${{\upsilon }}$ is frequency
\[\phi \] is work function
KE is the maximum kinetic energy.

Complete step by step Answer:
We have given that.
Frequency of radiation ${{\upsilon }}{{ = 1}}{{.4 \times 1}}{{{0}}^{{{15}}}}{{{H}}_{{z}}}$
Work function $\phi = {{2}}{{.8eV}}$
As the electron is accelerating with the anode voltage ${{3V}}$ its minimum kinetic energy will become $0$ (By Indent radiation)
We know that
${{h\upsilon = }}{{f + K}}{{{E}}_{{{max}}}}\_\_\left( 1 \right)$
Here
${{K}}{{{E}}_{{{min}}}}{{ = 0}} \to $ By Incident Radiation
Here the electrons are accelerated by potential ${{3V}}$, So energy we get from this is
So,
Energy due to accelerating voltage is equal to
${{E = qV}}$
${{ = q}}\left( {{{3V}}} \right)$ Here the charge on electron is ${{e}}$ so,
${{E = }}{{e}}\left( {{{3V}}} \right){{or}}{{E = }}{{3eV}}$
This energy will carry by all the electrons out here by accelerating potential
So, Kinetic Energy (minimum) of photoelectron is equal to
${KE_{min}}$ = ${0 + 3V}$
 $\Rightarrow {KE_{min}}$ = ${3eV}$
Now, calculating, Kinetic Energy (maximum) of photoelectron is equal to ${KE_{max}}$ + $3eV$
So, finding it with ${{Eq\_\_}}\left( {{1}} \right)$
${{or}}$ ${{h\upsilon - }}\phi$ = ${KE_{max}}$
$ \Rightarrow \dfrac{{{{6}}{{.625 \times 1}}{{{0}}^{{{ - 34}}}}}}{{{{1}}{{.6 \times 1}}{{{0}}^{{{ - 19}}}}}}{{ \times 1}}{{.4 \times 1}}{{{0}}^{{{15}}}}{{ - 2}}{{.8ev}} = {KE_{max}}$
$ \Rightarrow {{5}}{{.8 - 2}}{{.8}}$ = ${KE_{max}}$
$ \Rightarrow {{3eV}}$ = ${KE_{max}}$
So, final kinetic energy (maximum) is given by
${KE_{max}}$ = ${{3eV + 3eV}}$ (Because of anode voltage)
$ \Rightarrow {KE_{max}}$ = ${{3eV + 3eV}}$
$ \Rightarrow {KE_{max}}$ = ${{6eV}}$
So,
Minimum (Kinetic Energy) ${{ = }}{{3eV}}$
Maximum (Kinetic Energy) ${{ = }}{{6eV}}$

So, the correct option is (A). i.e. $\left( {3,6} \right)$.

Note: Here the electron has applied an extra anode voltage, so because of that extra accelerating voltage extra energy i.e. $\left( 3 \right)$ will be occupied by each electron contributing ${{3W}}$ extra kinetic energy to it. Photoelectric emission takes place only when threshold frequency is reached.