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Question: The surface of metal of work function \( \phi \) is illuminated by light whose electric field compon...

The surface of metal of work function ϕ\phi is illuminated by light whose electric field component varies with time as E=E0[1+cosωt]sinω0t{\rm E} = {{\rm E}_0}\left[ {1 + \cos \omega t} \right]\sin {\omega _0}t . Find the maximum kinetic energy of photoelectrons emitted from the surface.

Explanation

Solution

Hint : According to the Einstein photoelectric equation total energy of photon is equal to the sum of maximum kinetic energy and work function of the metal surface.
Mathematical formulation is given as-
Energy of photon == maximum kinetic energy of electron + Work function of metal
hν=12mv2+hν0h\nu = \dfrac{1}{2}m{v^2} + h{\nu _0} (1)(1)
Here in this question we have used formula for Energy of photon == maximum kinetic energy of electron + Work function of metal
\Rightarrow hν=12mv2+hν0h\nu = \dfrac{1}{2}m{v^2} + h{\nu _0} .
And cosαsinβ=12[sin(αβ)+sin(α+β)]\cos \alpha \sin \beta = \dfrac{1}{2}\left[ {\sin \left( {\alpha - \beta } \right) + \sin \left( {\alpha + \beta } \right)} \right]

Complete Step By Step Answer:
We are given that E=E0[1+cosωt]sinω0t{\rm E} = {{\rm E}_0}\left[ {1 + \cos \omega t} \right]\sin {\omega _0}t
Therefore the equation of given electric field component is, E=E0sinω0tcosωt=E0sinω0t+E02[sin(ω0+ω)t+sin(ω0ω)t]{\rm E} = {{\rm E}_0}\sin {\omega _0}t\cos \omega t = {{\rm E}_0}\sin {\omega _0}t + \dfrac{{{{\rm E}_0}}}{2}\left[ {\sin \left( {{\omega _0} + \omega } \right)t + \sin \left( {{\omega _0} - \omega } \right)t} \right] (2)(2)
Therefore, the given light consist of three different frequencies that are ω\omega , ω0+ω{\omega _0} + \omega , ω0ω{\omega _0} - \omega
The maximum kinetic energy will be due to most energetic photon
Now by using equation (1)(1) we have,
KEmax=hνϕK{E_{\max }} = h\nu - \phi
Here, h= Planck’s constant, ν\nu = frequency of light and ϕ=hν0\phi = h{\nu _0} = work function of the metal
Therefore equation we have,
KEmaxK{E_{\max }} =h(ω+ω0)2Πϕ= \dfrac{{h\left( {\omega + {\omega _0}} \right)}}{{2\Pi }} - \phi [Since ω=2Πv\omega = 2\Pi v or v=ω2Πv = \dfrac{\omega }{{2\Pi }} ]

Note :
We know that the Photoelectric effect was described by Einstein. He explained that when light rays are allowed to fall on a metal surface, electrons emitted from the surface are called photoelectrons and the effect is known as photoelectric effect. Energy of the photon is given by E=hνE = h\nu where, h is known as Planck’s constant and ν\nu is the frequency of light. Also photoelectric equation is given by
Energy of photon == maximum kinetic energy of electron + Work function of metal