Dual Nature of Radiation and Matter

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Questions:

Generic 5

7 x 5 = 35

The main aim of Davisson and Germer was to study about nickel surface by directing beam of electrons at its surface and note the number of electrons that bounced off at different angles. The carried out their experiment inside a vacuum chamber where an air after entering the chamber gives an oxide film on nickel surface. Again the does their experiment and found that the electrons which hits the nickel surface were scattered by atoms that appears from crystal planes in nickel crystal. By doing their regular experiment, Davisson and Germer's at once found diffraction of electrons which was an initial proof that confirm about de Broglie's hypothesis and shows wave properties in particles.
(a) What can we infer about the values shown by Davisson and Germer?
(b) Write the expression to find the wavelength of an electron when accelerated through a potential difference of V volts.

Ram knows that red light has greater and so it is much bright, but in case of photoelectric emission it cannot produce the emission of electrons from a clean zinc surface, while even weak ultraviolet radiation can do so. He could not know specific cause of such thing. Then he went to his friend Shyam for its specific explanation. Shyam explained him that the photoemission of electron does not depend on the intensity while it depends on the frequency and thus on the energy of photon of incident light. The energy of photon of red light cannot emit photoelectrons. Similarly, the energy of photon of ultraviolet light is greater than the work function of zinc, so ultraviolet light can emit photoelectrons.
(a) What values are noticed in Shyam?
(b) The work functions of lithium and copper are 2.3eV and 4eV respectively. Which of these metals are useful for the photoelectric cell working with visible light? Explain.

Mohan thought that there are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. But, can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength? He got confused and could not find its answer. Then he requested his friend Sohan. Sohan explained him that in the first case, energy given out is less than the energy supplied. But in the second case, the material has to supply the energy as the emitted photon has to supply the energy as the emitted photon has more energy, which cannot happen for stable substances.
(a) What values do you notice in Sohan?
(b) Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Radiation has dual nature,i.e., it possesses the properties of both; wave and particle.This prompted de-Broglie to predict dual nature of moving material particles.Thus waves are associated with moving material particles which are called matter waves. The wavelength of matter wave is given by \(\lambda =\frac { h }{ mv } \), where m is the mass, v is the speed of the particle and h is Plank's constant. Read the above paragraph and answer the following questions;
(i) How was the wave nature of the electron established?
(ii) What are the de-Broglie wavelength associated with a particle (i) at rest (ii) moving with infinite speed?
(iii) What are the basic values displayed with this study?

The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is 3.4 eV. Determine the de-Broglie wavelength associated with it.
Mass of electron = \(9.1\times { 10 }^{ -31 }Kg;\)
Planck's constant = \(6.63\times { 10 }^{ -34 }Js\)

If 10% of the energy supplied to an incandescent light bulb is radiated as visible light, how many visible light photons are emitted by 200-watt bulb? Assume wavelength of all visible photons to be \(5000\overset { \circ }{ A } \)Given \(h=6.63\times { 10 }^{ -34 }Js\)

A metal has a work function of 2.0 eV and is illuminated by monochromic light of wavelength 500nm. Calculate
(a) the threshold wavelength
(b) the maximum energy of photoelectrons
(c) the stopping potential.
\([use\ h=6.63\times { 10 }^{ -34 }Js;c=3\times { 10 }^{ 8 }{ ms }^{ -1 }]\)

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Answers Key:

Generic 5

7 x 5 = 35

(a) Perseverance, not giving up and patience.
(b) \(\lambda =\frac { 12.27 }{ \sqrt { V } } \)\(\overset { o }{ A } \)

(a) The values noticed in Shyam are:
(i) High degree of general awareness.
(ii) Concern for his friend.
(iii) Helping and caring nature.
(b) The threshold wavelength, \({ \lambda }_{ 0 }=\frac { hc }{ W } \)
For lithium, \({ \lambda }_{ 0 }=\frac { 12375 }{ 2.3 } \overset { 0 }{ A } =5380\overset { 0 }{ A } \)
For copper, \({ \lambda }_{ 0 }=\frac { 12375 }{ 4 } \overset { 0 }{ A } =3094\overset { 0 }{ A } \)
The wavelength 5380\(\overset { 0 }{ A } \) lies in visible region, thus lithium will be useful for photoelectric cell.

(a) The value noticed in Sohan are:
(i) High degree of general awareness.
(ii) Concern for his friend.
(iii) Helping and caring nature.
(b) Maximum energy \(=hv-\phi o\)
\(\Rightarrow \quad \left( \frac { 1230 }{ 600 } -\phi o \right) =\frac { 1 }{ 2 } \left( \frac { 1230 }{ 400 } -\phi o \right) \)
\(\Rightarrow \frac { 1230 }{ 300 } -2\phi o=\frac { 1230 }{ 400 } -\phi o=\frac { 1230 }{ 1200 } \)
= 1.02 eV

(i) Davisson and Germer observed diffraction patterns of slow moving electrons. And G.P. Thomson observed a diffraction pattern of fast moving electrons. As diffraction is essentially a wave phenomenon, therefore, it was concluded that wave must be associated with moving electrons.
(ii) (a) At rest, v = 0, \(=\frac { h }{ mv } =\frac { h }{ m\times 0 } =\infty \)
(b) Particle moving with infinite speed, v = \(\infty \)
\(\lambda =\frac { h }{ mv } =\frac { h }{ m\times \infty } \)
(iii) The dual nature of moving material particles reveals in a way the nature of Almighty God. He is in a visible form (i.e., Sakar) like a visible particle and also without any form (i.e., Nirakar) like a wave.It depends on us how we realize him.

\(Here,K.E. \ of \ electron,\)
\( k=3.4eV=3.4\times 1.6\times { 10 }^{ -19 }J\)
de-Broglie wavelength associated with electron
\(\lambda =\frac { h }{ mv } =\frac { h }{ \sqrt { 2mK } }\)
\(=\frac { 6.63\times { 10 }^{ -34 } }{ \sqrt { 2\times \left( 9.1\times { 10 }^{ -31 } \right) } \times 3.4\times 1.6\times { 10 }^{ -19 } } =6.63\times { 10 }^{ -10 }m\)

\(Here,\lambda =5000\overset { \circ }{ A } =5\times { 10 }^{ -7 }m;\)
Energy of one photon,
\(E=\frac { hc }{ \lambda } =\frac { \left( 6.63\times { 10 }^{ -34 } \right) \left( 3\times { 10 }^{ 8 } \right) }{ 5\times { 10 }^{ -7 } } =3.96\times { 10 }^{ -19 }J\)
A 200 W bulb supplies 200 J of energy per second.Energy emitted by lamp per second as visible light,
\({ E }_{ 1 }=200\times \frac { 10 }{ 100 } =20J{ s }^{ -1 }\)
Number of photons emitted per second as visible light.
\(N=\frac { { E }_{ 1 } }{ E } =\frac { 20 }{ 3.96\times { 10 }^{ -19 } } =5.05\times { 10 }^{ 19 }\)

\((a) \ Here,{ \phi }_{ 0 }=2.0eV\)
\( =2.0\times 1.6\times { 10 }^{ -19 }J;\lambda =500\times { 10 }^{ -9 }m\)
\( As, \ \ { \phi }_{ 0 }=\frac { hc }{ { \lambda }_{ 0 } }\)
\( or \ { \lambda }_{ 0 }=\frac { hc }{ { \phi }_{ 0 } } =\frac { \left( 6.63\times { 10 }^{ -34 } \right) \times \left( 3\times { 10 }^{ 8 } \right) }{ 2.0\times 1.6\times { 10 }^{ -19 } } =6.1875\times { 10 }^{ -7 }m=6187.5\overset { \circ }{ A } \)
(b) Max, kinetic energy of photoelectron is
\({ K }_{ max }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\(or \ \ =\frac { \left( 6.63\times { 10 }^{ -34 } \right) \times \left( 3\times { 10 }^{ 8 } \right) }{ \left( 500\times { 10 }^{ -19 } \right) \times \left( 1.6\times { 10 }^{ -19 } \right) } -2.0\)
\(=2.475-2.0=0.475eV\)
(c) Stopping potential
\({ V }_{ 0 }=\frac { { K }_{ max } }{ e } =\frac { 0.475eV }{ e } =0.475V\)