Question

Electromagnetic radiations described by the equation $E=100\left[ Sin(2\times{{10}^{15}})t+Sin(6.28\times{{10}^{15}})t \right]\dfrac{V}{m}$ are incident on photosensitive surface having work function $1.5eV$. The maximum kinetic energy of the photoelectron will be.
A. $0.125eV$
B. $2.625eV$
C. $4.125eV$
D. $6.125eV$

Answer 1

Electromagnetic waves are also known as EM waves that are produced when an electric field comes in
contact with the magnetic field. It can also be said that electromagnetic waves are the composition of oscillating electric and magnetic fields. Electromagnetic waves are solutions of Maxwell’s equations, which are the fundamental equations of electrodynamics.

Complete step by step solution:
(1). As we know electromagnetic radiation standard equation is
$E={{E}_{o}}\left[ Sin({{w}_{1}}t)+Sin({{w}_{2}}t) \right]$

(2). Then find frequency ${{v}_{1}}$ and ${{v}_{2}}$

(3). For maximum kinetic energy
$K{{E}_{moy}}=E-\phi$
Where $E\Rightarrow$ Energy $[E=hv]$
$\phi \Rightarrow$work function
As the work function is fixed, to maximum KE we have to increase energy so we take maximum frequency.
Generally, an electric field is produced by a charged particle. A force is exerted by this electric field on other charged particles. Positive charges accelerate in the direction of the field and negative charges accelerate in a direction opposite to the direction of the field.
The Magnetic field is produced by a moving charged particle. A force is exerted by this magnetic field on other moving particles. The force on these charges is always perpendicular to the direction of their velocity and therefore only changes the direction of the velocity, not the speed.

Given that the equation is
$E=100\left[ \sin (2\times {{10}^{15}})t+\sin (6.28\times {{10}^{15}})t \right]v/m$
Compare the equation with
$E={{E}_{0}}\left[ \sin ({{w}_{1}}t)+\sin ({{w}_{2}}t) \right]$
So, ${{w}_{1}}=2\times {{10}^{15}}$ red/sec
${{w}_{2}}=6.28\times {{10}^{15}}$red/sec
We know $w=2\pi v$
So,${{v}_{1}}=\dfrac{{{w}_{1}}}{2\pi }$
$=\dfrac{2\times {{10}^{15}}}{6.28}Hz$
${{v}_{2}}=\dfrac{{{w}_{2}}}{2\pi }=\dfrac{6.28\times {{10}^{15}}}{6.28}={{10}^{15}}Hz$
By Comparison
${{v}_{2}}>{{v}_{1}}$
So for max energy we use ${{v}_{2}}$
Energy for ${{v}_{2}}$ photon, $E=h{{v}_{2}}$
$=6.6\times {{10}^{-34}}\times {{10}^{15}}$
$=6.6\times {{10}^{-19}}$Jaule
Lev $=6.6\times {{10}^{-19}}$ J
So $E=\dfrac{6.6\times {{10}^{-19}}Jaule}{1.6\times {{10}^{-19}}J/ev}$
$=4.125ev$
So maximum kinetic energy
KE may $=E-\phi$
$=4.125-1.5$
$=2.625ev$

Hence, Option (B) is the correct answer.

Note: Students know the standard electromagnetic radiation equation.
Know the relation between kinetic energy and work function.
Kinetic energy of photoelectron measures in eV.