Question

The magnetic field associated with a light wave is given, at the origin, by $ B = {B_o}\left[ {\sin \left( {3.14 \times {{10}^7}} \right)ct + \sin \left( {6.28 \times {{10}^7}} \right)ct} \right] $ . If this light falls on a silver plate having a work function of $ 4.7eV $ , what will be the maximum kinetic energy of the photo electrons?
 $ \left( {c = 3 \times {{10}^8}m{s^{ - 1}};h = 6.6 \times {{10}^{ - 34}}J - s} \right) $
(A) $ 7.72eV $
(B) $ 8.52eV $
(C) $ 12.5eV $
(D) $ 6.82eV $

Answer 1

Hint: As the magnetic field is made up of 2 waves we need to find the frequency of the wave which is the maximum. Then using that frequency, we need to find the energy of the photons. The energy of the photo electrons will be the energy of the photons minus the work function of the metal.

Formula Used: In this solution we will be using the following formulas,
 $ \Rightarrow K.E. = E + \phi $
where $ K.E. $ is the kinetic energy of the photo electrons,
 $ E $ is the energy of the photons and
 $ \phi $ is the work function of the metal.

Complete step by step answer
In the question, we are given the magnetic field associated with the light wave. Here we can see that the magnetic field consists of two waves. So the photons having the maximum energy will be coming from the wave having the maximum frequency. In the question the two waves given are,
 $ \Rightarrow {B_1} = {B_o}\sin \left( {3.14 \times {{10}^7}c} \right)t $
The amount $ 3.14 \times {10^7}c $ is the frequency of the wave and let us denote it by $ {\omega _1} = 3.14 \times {10^7}c $
The second wave is
 $ \Rightarrow {B_2} = {B_o}\sin \left( {6.28 \times {{10}^7}c} \right)t $ . In this case similarly the frequency of the wave will be, $ {\omega _2} = 6.28 \times {10^7}c $
So the frequency of the photons due to these waves will be $ {\upsilon _1} $ and $ {\upsilon _2} $ and their values will be given by
 $ \Rightarrow {\upsilon _1} = \dfrac{{{\omega _1}}}{{2\pi }} $ and $ {\upsilon _2} = \dfrac{{{\omega _2}}}{{2\pi }} $
Therefore, substituting the values we get,
 $ \Rightarrow {\upsilon _1} = \dfrac{{3.14 \times {{10}^7}c}}{{2\pi }} $
So we get,
 $ \Rightarrow {\upsilon _1} = \dfrac{1}{2} \times {10^7}c $ and,
 $ \Rightarrow {\upsilon _2} = \dfrac{{6.28 \times {{10}^7}c}}{{2\pi }} $
So we get,
 $ \Rightarrow {\upsilon _2} = {10^7}c $
Hence we can see that the value of $ {\upsilon _2} $ is more than $ {\upsilon _1} $ . So the frequency of the photons will be $ {\upsilon _2} $ . From the frequency of the photons, we can get their energy as,
 $ \Rightarrow E = h{\upsilon _2} $
In the question we are given $ c = 3 \times {10^8}m{s^{ - 1}} $ and $ h = 6.6 \times {10^{ - 34}}J - s $ .
So substituting the values we get,
 $ \Rightarrow E = 6.6 \times {10^{ - 34}} \times {10^7} \times 3 \times {10^8} $
Calculating we get,
 $ \Rightarrow E = 1.98 \times {10^{ - 18}}J $
Now to convert this into electron volt, the conversion factor is,
 $ \Rightarrow 1J = 6.24 \times {10^{18}}eV $
So we get,
 $ \Rightarrow E = 1.98 \times {10^{ - 18}} \times 6.24 \times {10^{18}}eV $
This gives us,
 $ \Rightarrow E = 12.4eV $
Now, the kinetic energy of the photo electrons are given by
 $ \Rightarrow K.E. = E + \phi $
In the question we are given $ \phi = 4.7eV $
So substituting we get,
 $ \Rightarrow K.E. = \left( {12.4 - 4.7} \right)eV $
This is equal to, $ K.E. = 7.7eV \simeq 7.72eV $
So the kinetic energy of the photo electrons is equal to $ 7.72eV $ .
So the option (A) is correct.

Note
The photoelectric effect is a phenomenon where electrons are emitted from the surface of metals when light of sufficient frequency strikes on it. The kinetic energy of these electrons increases with the increase in the intensity of the light wave.