The Photon, the Quantum of Light MCQ Quiz - Objective Question with Answer for The Photon, the Quantum of Light - Download Free PDF

Concept:

Energy of a Photon: The energy of a photon is given by the equation:

Energy (E) = h × ν

h: Planck's constant (6.6 × 10^-34 J·s)

ν: Frequency of the wave

Frequency and Wavelength Relationship: The frequency (ν) and wavelength (λ) of the wave are related by the equation: ν = c / λ

c: Speed of light (3 × 10⁸ m/s)

λ: Wavelength

Power and Photons: The total power emitted is related to the energy per photon and the number of photons emitted per second:

Power (P) = Energy per photon × Number of photons per second

Calculation:

• Power of the bulb, P = 66 W
• Wavelength of the emitted wave, λ = 600 nm = 600 × 10^-9 m
• Planck's constant, h = 6.6 × 10^-34 J·s
• Speed of light, c = 3 × 10⁸ m/s

The frequency (ν) can be calculated using:

ν = c / λ

Substituting the given values:

ν = (3 × 10⁸ m/s) / (600 × 10^-9 m) = 5 × 10¹⁴ Hz

The energy of one photon is:

Energy (E) = h × ν = (6.6 × 10^-34 J·s) × (5 × 10¹⁴ Hz) = 3.3 × 10^-19 J

The number of photons emitted per second is:

Number of photons = Power / Energy per photon = 66 W / (3.3 × 10^-19 J)

Number of photons = 2 × 10²⁰ photons per second

∴ The number of photons emitted per second by the bulb is 2 × 10²⁰ photons.

Concept:

de Broglie Wavelength: The de Broglie wavelength of a particle is given by the formula:

λ = h / p, where:

λ: de Broglie wavelength

h: Planck's constant (6.626 × 10^-34 J·s)

p: Momentum of the particle

For an electron accelerated through a potential difference, the kinetic energy is given by eV = (1/2)mv².

The momentum of the electron can be found from its kinetic energy: p = √(2m × e × V).

The de Broglie wavelength is then calculated using λ = h / √(2m × e × V).

Calculation:

Given:

• Potential difference, V = 64 V
• Electron mass, m = 9.11 × 10^-31 kg
• Electron charge, e = 1.6 × 10^-19 C
• Planck's constant, h = 6.626 × 10^-34 J·s

Substitute the values into the de Broglie wavelength formula:

λ = (6.626 × 10^-34) / √(2 × 9.11 × 10^-31 × 1.6 × 10^-19 × 64)

λ ≈ 1.43 Å

∴ The de Broglie wavelength associated with the electron is 1.43 Å.

Concept Used:

de-Broglie Wavelength:

The de-Broglie wavelength (λ) associated with a moving particle is given by the formula:

λ = h / (m × v)

Where:

λ = de-Broglie wavelength (m)

h = Planck's constant = 6.63 × 10^-34 J·s

m = Mass of the particle (kg)

v = Speed of the particle (m/s)

SI Unit of wavelength: Meter (m)

Calculation:

Given,

Mass of electron, m = 9.11 × 10^-31 kg

Speed of electron, v = 6.4 × 10⁶ m/s

Planck's constant, h = 6.63 × 10^-34 J·s

Using the de-Broglie wavelength formula:

λ = h / (m × v)

⇒ λ = (6.63 × 10^-34) / (9.11 × 10^-31 × 6.4 × 10⁶)

⇒ λ ≈ 0.114 nm

∴ The de-Broglie wavelength associated with the electron is 0.114 nm.

Concept Used:

The number of photons emitted per second by a light source is determined using the energy of a photon and the total power of the beam.

The energy of a single photon is given by:

E = h c / λ

where:

h = Planck’s constant = 6.6 × 10^-34 J·s

c = Speed of light = 3 × 10⁸ m/s

λ = Wavelength of light = 667 × 10^-9 m

The number of photons emitted per second (N) is given by:

N = P / E

where P is the total power emitted by the laser.

Calculation:

Given:

P = 9 × 10^-3 W

Substituting the values:

N = (9 × 10^-3) × (667 × 10^-9) / [(6.6 × 10^-34) × (3 × 10⁸)]

N = 3 × 10¹⁶ photons/sec

∴ The number of photons arriving per second is 3 × 10¹⁶.

\(\lambda = \dfrac{ 6.6 \times 10 ^ {-34} }{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 50}} = 1.72 \times 10^{-10} = 1.72 A\)

Option B

CONCEPT:

• Light: It is an electromagnetic wave.
  • Light shows dual nature i.e. wave and particle.

Phenomenon explaining the wave nature of light is/are:

1. Interference
2. Diffraction
3. Polarisation

Phenomenons explaining the particle nature of light are:

1. Photoelectric effect
2. Compton Scattering

• Photons associated with light shows that the energy is Quantised and occurs in discrete levels.

EXPLANATION:

• From the above discussion, the dual nature of light is exhibited by diffraction and the photoelectric effect . So option 1 is correct.

CONCEPT:

Light shows dual nature:

• Some phenomena can be explained by the wave nature of light while some by the quantum nature of light.

EXPLANATION:

Photoelectric effect: 

• The emission of free electrons from a metal surface when the light is illuminated on it, it is called the photoemission or the photoelectric effect. 
• This effect led to the conclusion that light is made up of packets or quantum of energy.

Diffraction: 

• Diffraction is the bending of light at the sharp ends of the obstacles or holes. It can be explained by the wave nature of light.

Polarization of light:

• The light waves whose vibrations occur in a single plane are called polarized light.
• The light wave whose vibrations occur in more than one plane is called un-polarized light.
• The phenomenon of the transformation of un-polarized light into polarized light is called polarization of light.

• Polarization, in Physics, is defined as a phenomenon caused due to the wave nature of electromagnetic radiation. Hence, option 1 is correct.

Concept:

Corpuscle nature of light:

• The light is made up of small discrete particles named as "corpuscles" (little particles) which travel in a straight line and have a finite velocity. 
• It basically says about the particle nature of light. 

Compton Effect:

• It is the scattering of a photon by a charged particle, usually an electron.
• When a photon collides with an electron at rest, the photon gives its energy to the electron.
• So, the scattered photon (photon after the collision) will have a higher wavelength (lower energy) compared to the wavelength of the incident photon (photon before the collision).
• This shift in wavelength is called the Compton shift.

​Explanation:

So, it explains the collision of photons with electrons. 

The energy of the photon got changed due to a change in wavelength after the collision. This is the behaviour of the particle or Corpuscle nature of light. 

So, the correct option is Corpuscle nature of light.

​Additional Information

• The Compton effect is named after American physicist Arthur Holly Compton, who first discovered this effect. 
• The photoelectric effect also shows particle nature.
• The wave nature of light is shown by Hygueins Princile.
• The transverse nature of light is shown by Polarization. 

CONCEPT:

Photon:

According to Einstein's quantum theory light propagates in the bundles (packets or quanta) of energy, each bundle is called a photon and possessing energy.

The energy of photon:

The energy of each photon is given by

\(⇒ E = h\nu = \frac{{hc}}{λ }\)
Where c = Speed of light, h = Plank's constant = 6.6 × 10–34 J-sec, ν = Frequency in Hz, λ  = Wavelength of light

The momentum of the photon:

\(⇒ p = m \times c = \frac{E}{c} = \frac{{h\nu }}{c} = \frac{h}{λ }\)

EXPLANATION:

For the same energy, the momentum of the electron is more than that of the photon, i.e.,

⇒ p_e > p_ph

The wavelength of the electron is 

\(⇒ λ_e = \frac{h}{p_e }\)

The wavelength of the photon is 

\(⇒ λ_{ph} = \frac{h}{p_{ph} }\)

As pe > p_ph therefore, λph > λe

CONCEPT:

Cathode rays: 

• Cathode rays are the stream of high-speed negatively charged particles moving from cathode to anode in a discharge tube.

Properties of Cathode Rays:

• Cathode rays travel in a straight line.
• They are streams of fast-moving electrons.
• Cathode rays heat up the material on which they fall.
• The cathode rays in the discharge tube are the electrons produced due to the ionization of gas and that emitted by the cathode due to the collision of positive ions.
• Cathode rays are emitted normally from the cathode surface. Their direction is independent of the position of the anode. 
• Cathode rays are deflected by an electric field and also by a magnetic field.
• Cathode rays ionize the gases through which they are passed.
• Cathode rays can penetrate through the thin foils of metal.

Explanation:

A moving photon has the mass that is given by:

\(E = mc^2\)

\(m = \frac{E}{c^2}\)     ------(1)

We know that the energy of the photon is 

\(E=h\nu\)     ------(2)

Combining (1) and (2)

\(m = \frac{E}{c^2}\)

\(m = \frac{h\nu}{c^2}\)

CONCEPT:

• The photon is a type of elementary particle. It is a basic unit of light. Photons are massless, and they always move at the speed of light in a vacuum ( c = 3 × 108  m/s)
• Photon energy is the energy that is carried by a single photon. The amount of this energy is inversely proportional to the wavelength.
  • If the wavelength increases, energy will be lowered.

The equation of photon energy is given by:

\(E=\frac{hc}{λ}\)

where E is energy, h is the Planck constant, c is the speed of light in a vacuum and λ is the photon's wavelength. In this equation h and c both are constants.

CALCULATION: 

Given that λ = 2000 Å = 2000 × 10-10 m

h = 6.626 × 10-34  (constant)

c =  3 × 108   (constant)

E = (h × c) / λ 

E = 6.626 × 10-34 × 3 × 108 / 2000 × 10-10

E = 9.94 × 10-19 J

The correct answer is Option 4 i.e. 9.94 × 10-19 J

Concept:

• The wavelength of the electron due to its motion is called the de-Broglie wavelength of the electron.
• The bundle of light rays is called a photon.
• The de-Broglie wavelength of an electron (λe) is given by:

\(\Rightarrow {λ _e} = \frac{h}{mv}= \frac{12.27}{\sqrt{V}}A^{\circ}\)

• Energy of a photon (E) = (hc)/λ

∴ The wavelength of the photon is

\(⇒ λ =\frac{{h\;c}}{E}\)

Where E = energy, h = Planck constant, m = mass of electron and c = speed of light

Explanation:

• The wavelength associated with an electron accelerated through a potential difference of V is 

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

\(\Rightarrow {λ _e} = \frac{h}{mv}= \frac{12.27}{\sqrt{V}}A^{\circ}\)

CONCEPT:

• Electric bulb: It is an electric device that converts electric energy into heat and light energy.
  • it is generally made of glass with inert gases filled in it (Nitrogen and argon).
  • The filament is made of tungsten which has a very high melting point.
• Photon: The bundle of light is called a photon.

EXPLANATION:

• When an electric current flows in a bulb, the filament of the bulb is heated and emits photons.
• The photon released by the filament lies in the visible range of the electromagnetic spectrum.
• We say this emitting of the photon as light. Hence option 2 is correct.
• The particles like- electrons, protons, and alpha are never released by the filament. 

CONCEPT:

• The photon is a type of elementary particle. It is a basic unit of light. 
• Photons are massless, and they always move at the speed of light in a vacuum (c = 3 × 108  m/s).
• Photon energy is the energy that is carried by a single photon. The amount of this energy is inversely proportional to the wavelength.
• If the wavelength increases, energy will be lowered.
• The equation of photon energy is 

​\(E=\frac{hc}{λ}\)

where E is energy, h is the Planck constant, c is the speed of light in vacuum and λ is the photon's wavelength.

• The amount of input energy converted into output is known as Efficiency. 

CALCULATION:

Given - Power (P) = 200 W and  wavelength (λ) = 0.6 μm 

The amount of energy converted into the  light is 25 %

• The energy of a photon in a yellow light is  given by

\(\Rightarrow E = \frac{hc}{\lambda}=\frac{6.626× 10^{-34}× 3 ×10^8}{0.6×10^{-6}}=33×10^{-20}J\)

• The energy radiated in the form of light

\(\Rightarrow 200 ×\frac{25}{100}=50 watt\)

• The number of photons emitted per seconds is 

\(\Rightarrow\frac{Total \, power}{Energy \,of\,each\, photon }=\frac{50}{33×10^{-20}}=1.5×10^{20}\)