Dual Nature of Matter and Radiation Test - 1

• The photoelectric effect is based on the particle nature of radiation since it is a phenomenon in which the interaction of particle takes place with the matter.
• The photoelectric effect is the phenomenon when electrons are ejected from the surface of a metal when light of sufficient frequency falls on it.

CONCEPT:

Louis de Broglie in his theory on wave nature of matter proposed that:

All particles could be treated as matter waves with a wavelength $λ.$ And their frequency is given by the following equation:

\(λ = \frac{h}{mv}\)

where λ is de Broglie wavelength, h is Planck's const, m is mass, v is the velocity.

Calculation:

Given:

λ = 10-10 m

From equation 1);

\({10^{ - 10}} = \frac{{6.6\; \times \;{{10}^{ - 34}}}}{{9.1\; \times \;{{10}^{ - 31}}\; \times \;v}};\)

\(v = \frac{{6.6\; \times \;{{10}^{ - 34}}}}{{9.1\; \times \;{{10}^{ - 41}}}} \)

v = 7.25 x 10^-6 m/s

CONCEPT:

Photoelectric effect:

• When the light of a sufficiently small wavelength is incident on the metal surface, electrons are ejected from the metal instantly. This phenomenon is called the photoelectric effect.

Stopping potential:

• It is defined as the potential required to stop the ejection of an electron from a metal surface when the incident beam of energy is greater than the work function of the metal.

Work function:

• It is the minimum amount of energy required so that metal emits an electron. It is represented by ϕ. ​

Photon:

• According to Einstein's theory, the light propagates in the bundles of energy, each bundle is being called a photon.
• ​The rest mass of the photon is zero. But its effective mass is given as,
• Photon exerts pressure on the surface.
• The energy of a photon is given as,

​\(⇒ E=\frac{hc}{λ}\)

Where h = Planck's constant, c = speed of light in vacuum and λ = wavelength​

• We know that the energy of one photon is given as,

\(⇒ E=\frac{hc}{λ}\)

\(⇒ E=\frac{6.6×10^{-34}×3×10^8}{6000×10^{10}}\)

⇒ E = 3.3 × 10^-19 J

• If n number of the electrons are emitted per second, then,

⇒ ηP = nE

\(\Rightarrow n=\frac{ηP}{E}\)

\(⇒ n=\frac{4\times20}{3.3×10^{-19}\times100}\)

⇒ n = 24.2 × 1017

• Hence, option 4 is correct.

Concept:

Quantum nature of light:

• Light consists of photons or quanta of energy that gives particle nature to it.
• The Photoelectric effect can be explained by the quantum nature of light.
• A photon is an elementary particle. It is a quantum of light.
• The photon is a chargeless (neutral) particle.
• The energy of a photon is given as,

\(\Rightarrow E=hν=\frac{hc}{λ}\)

Where h = Planck's constant, ν = frequency of light, c = speed of light, and λ = wavelength of light

Calculation:

Given:

ν = 14 GHz = 14 × 10⁹ Hz, h = 6.626 x 10-34 J s

Using, E = hν 

E = 6.626 x 10-34 × 14 × 109 

E = 9.28 × 10-24 J

CONCEPT:

Photon:

• According to Einstein's theory, the light propagates in the bundles of energy, each bundle is being called a photon.
• ​The rest mass of the photon is zero. But its effective mass is given as,
• Photon exerts pressure on the surface.
• The energy of a photon is given as,

\(⇒ E=hν=\frac{hc}{λ}\)

Where h = 6.63 × 10-34 J-sec = Planck's constant, c = 3 × 108 m/sec = speed of light, ν = frequency, and λ = wavelength

Particle nature of light: The Photon

• The photoelectric effect gave evidence to the strange fact that light in interaction with matter behaved as if it was made of quanta or packets of energy.
• Einstein arrived at the important result, that the light quantum can also be associated with momentum.
• The momentum of the light quantum is given as,

\(⇒ P=\frac{hν}{c}\)

• A definite value of energy, as well as momentum, is a strong sign that the light quantum can be associated with a particle. That particle was later named photon.
• The particle-like behavior of light was further confirmed, in 1924, by the experiment of A.H. Compton (1892-1962) on the scattering of X-rays from electrons.
• We can summarise the photon picture of electromagnetic radiation as follows:
  1. In the interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
  2. Each photon has energy E (= hν) and momentum p (= hν/c), and speed c, the speed of light.
  3. All photons of light of a particular frequency or wavelength have the same energy and momentum, whatever the intensity of the radiation may be. By increasing the intensity of light of a given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of the intensity of radiation.
  4. Photons are electrically neutral and are not deflected by electric and magnetic fields.
  5. In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, the number of photons may not be conserved in a collision. The photon may be absorbed or a new photon may be created.

CALCULATION:

Given E = 1.326 × 10-19 J

• We know that the energy of a photon is given as,

⇒ E = hν     ---(1)

Where h = 6.63 × 10-34 J-sec = Planck's constant, and ν = frequency

By equation 1,

\(⇒ ν=\frac{E}{h}\)

\(⇒ ν=\frac{1.326\ ×\ 10^{-19}}{6.63\ ×\ 10^{-34}}\)

⇒ ν = 2 × 10¹⁴ Hz

Hence, option 1 is correct.

Concept:

• The wavelength of an electron due to its motion is called the de-Broglie wavelength of the electron.

The de-Broglie wavelength of the electron (λe) is given by:

λe = h/p = h/(mv)

\({λ _e} = \;\frac{h}{{\sqrt {2m\;\left( {KE} \right)} }}\)

Where, h is Planck's constant, P is momentum, v is velocity, m is mass of the particle, KE is the kinetic energy of the particle.

Calculation:

Given that,

Kinetic energy of electron = 120 eV

Now according to De Broglie waves equation

The wavelength of matter-wave, \(\lambda = \frac{h}{p}\)

And Kinetic energy of the particle is given as \(K.E = \frac{{{p^2}}}{{2m}}\)

Therefore \(p = \sqrt {2mE} \)

\( \sqrt {2 \times 9 \times {{10}^{ - 31}} \times 120 \times 1.6 \times {{10}^{ - 19}}} = \;5.88 \times {10^{ - 24}}\;kg\;m/s\)

Hence \(\lambda = \frac{h}{{\sqrt {2mE} }} = \frac{{6.63 \times {{10}^{ - 34}}}}{{5.88 \times {{10}^{ - 24}}}} \)

= 0.1120 x 10^-9 m = 112 pm

CONCEPT:

Dual nature of matter:

• According to de Broglie, the matter has a dual nature of wave-particle.
• So the matter is said to possess dual nature, i.e., it has both the properties of a particle and as well as a wave.
• The wave associated with each moving particle is called matter waves.
• ​​de Broglie wavelength associated with the particle is given as,

\(⇒ λ = \frac{h}{mv}=\frac{h}{P}\)   

Where, h = Planck's constant, m = mass of a particle, v = velocity of a particle and P = momentum

EXPLANATION:

• According to de Broglie when a particle moves, it will have waves of a definite wavelength associated with them.

According to de Broglie, the wavelength of matter waves is given as,

\(⇒ λ = \frac{h}{mv}=\frac{h}{P}\)     -----(1)   

Where, h = Planck's constant, m = mass of a particle, v = velocity of a particle and P = momentum

By equation 1 for a particle of mass m,

\(⇒ λ \propto \frac{1}{v}\)

• Since for a particle, the wavelength of its matter-wave is inversely proportional to its velocity, so we can say that the faster the particle moves, the shorter is the wavelength.

By equation 1 if velocity is the same for different particles,

\(⇒ λ \propto \frac{1}{m}\)

• Since when the velocity is the same for different particles, the wavelength of their matter-wave is inversely proportional to the mass of the particles, so we can say that for the same velocity, a heavier particle has a shorter wavelength.

By equation 1,

\(⇒ λ \propto \frac{1}{P}\)

• Since the wavelength of their matter-wave is inversely proportional to the momentum, so we can say that the higher the momentum, the shorter is the wavelength.
• Therefore the statement in option 2 is not true.
• Hence, option 2 is correct.

CONCEPT:

•  Photoelectric effect: It is the emission of electrons when electromagnetic radiation, such as light, hits a material.
  • Electrons emitted in this manner are called photoelectrons. 
• Photosensitivity: It is the amount to which an object reacts upon receiving photons, especially visible light.

Einstein's photoelectric equation:

E = hν

where h = Plank's constant ( 6.6261 × 10-34 Js.), ν = Frequency (Hz)

The Energy of the photon:

E = ϕ + KE 

where, ϕ = work function, K.E = Maximum kinetic energy of the electron.

CALCULATION:

Given, Photon energy E₁ = hν₁ = 1 eV, E₂ = 2.5 eV, W = 0.5 eV, 

Let v₁ and v₂ be the speed of the emitted electrons

Using Einstein's Photoelectric equation

K.E = E - ϕ

\(\Rightarrow \frac{1}{2}mv_1^2 = 1 - 0.5 \Rightarrow {v_1} = \sqrt {\frac{{2 \times 0.5}}{m}} \)

\(\Rightarrow \frac{1}{2}mv_2^2 = 2.5 - 0.5 \Rightarrow {v_2} = \sqrt {\frac{{2 \times 2}}{m}} \)

\(\Rightarrow \frac{{{v_1}}}{{{v_2}}} = \frac{1}{2}\)

Hence, the ratio of the maximum speeds of the emitted electrons is 1:2

CONCEPT:

• Photoelectric effect: When electromagnetic radiation hits a material, electrons are emitted, this effect is known as the Photoelectric effect.
  • Electrons emitted in this manner are called photo-electrons.

Max Kinetic energy of these photo-electrons are given by:

K.E.max = h f - ϕ 

where h is the Planck constant f is the frequency of the incident light or electromagnetic radiation. The term ϕ is the work function.

• The work function is the property of the material.

EXPLANATION:

From the graph slope is h.

The Kinetic energy of these photo-electrons is given by:

K.E. = h f - ϕ 

The slope of the above equation is h.

(in equation y = mx+c; slope is m)

And h is the plank const. which is a constant and independent of all variables.

So the correct answer is option 1.

CONCEPT:

Davisson and Germer Experiment:

• In 1927, and Germer designed an experiment to study the wave properties of electrons.
• Here the electrons emitted by the hot filament of an electron gun are accelerated by applying a suitable potential difference ‘V’ between the cathode and anode.
• The finely collimated beam of electrons from the electron gun is directed against the face of Ni crystal. The crystal is capable of rotation about an axis perpendicular to the plane of the paper.
• The electrons, scattered in different directions by the atoms of Ni crystal, are received by a movable detector which is just an electron collector.
• Thus we measure scattered electron intensity as a function of the scattering angle Φ, the angle between the incidence and the scattered electron beam. 

EXPLANATION:

• Electron gun emits electrons via thermionic emission.
• Two opposite charged plates are used to accelerate the electron called as electrostatic particle accelerators. Therefore statement 1 is correct.
• Nickel crystal used as target and electron beam is fired on Nickel crystal. Therefore statement 2 is correct.
• Detector is used to capture the scattered electron beam. Therefore statement 3 is correct.