Download CBSE Sample Paper 2024-25 for class 12th to 8th

Dual Nature of Matter and Radiation Test - 2

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Dual Nature of Matter and Radiation Test - 2

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Question 1
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Photoelectric effect occurs only if

A
Frequency of incident light is lesser than threshold frequency

B
Frequency of incident light is greater than threshold frequency

C
Frequency of incident light is equal to threshold frequency

D
Frequency of incident light is is equal to or greater than threshold frequency
Solution
The correct answer is option 2) i.e. Frequency of incident light is greater than the threshold frequency
CONCEPT:
- Photoelectric effect: The photoelectric effect is a phenomenon where electrons are ejected from a metal surface when the light of sufficient frequency is incident on it.
  When a photon falls on the metal surface, the photon’s energy is transferred to the electron.
  Some part of the energy gets utilized in removing the electron from the metal surface, and the remaining goes into giving kinetic energy for the ejected electron.
Therefore, the total energy of photon = work function + maximum kinetic energy of the electron.
- The energy of a photon is given by the equation:
$E =hν$
Where ν is the frequency of incident light and h is the Planck's constant.
- Work function: It is the minimum amount of energy required to cause photo-emission of electrons from a metal surface when light is incident on it.
- The work function is also known as the threshold energy.
EXPLANATION:
- The photoelectric effect is said to have occurred when an electron is ejected from the metal surface with light incident on it.
- For an electron to be emitted, it must acquire energy large enough to leave the metal surface. For this to happen, the electron should have an energy greater than the threshold energy.
- Since energy is directly proportional to frequency, the electron must acquire a frequency greater than the threshold frequency.
- Therefore, the frequency of incident light has to be greater than the threshold frequency for the photoelectric effect to occur.
Question 2
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By Huygens’s wave theory of light, we cannot explain the phenomenon of

A
Interference

B
Diffraction

C
Photoelectric effect

D
Polarisation
Solution
CONCEPT:
Huygen's Wave Theory:
- According to wave theory, a luminous body is a source of disturbance in a hypothetical medium ether. This medium pervades all space.
- It is assumed to be transparent and having zero inertia. The disturbance from the source is propagated in the form of waves through space.
- The waves carry energy and momentum. Huygen assumed that the waves were longitudinal.
- Further when polarization was discovered, then to explain it, light waves were, assumed to be transverse in nature by Fresnel.
- This theory explains successfully, the phenomenon of interference and diffraction apart from other properties of light.
- Huygen's theory fails to explain the photoelectric effect, Compton's effect, etc
- The wave theory introduces the concept of a wavefront.
EXPLANATION:
- Light shows dual nature i.e. wave and particle. Huygens’s theory is based on the wave nature of light.
- From above it is clear that Huygens’s wave theory of light, we cannot explain the phenomenon of Photoelectric effect. Therefore option 3 is correct.
Question 3
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If the energy of an electron becomes four times, then the de Broglie wavelength of the electron will become:

A
Double

B
Half

C
Remain Constant

D
None of these
Solution
CONCEPT:
De Broglie wave:
- According to de Broglie matter has a dual nature of wave-particle.
- The wave associated with each moving particle is called matter waves.
- Characteristics of Matter waves:
  The lighter the particle, the greater is the de Broglie wavelength.
  The higher the velocity of the particle, the smaller is its de Broglie wavelength.
  The de Broglie wavelength of a particle is independent of the charge or nature of the particle.
  The matter waves are not electromagnetic in nature. Only charged particles produce electromagnetic waves.
- de Broglie wavelength associated with the particle
$⇒ λ = \frac{h}{P}$
Where, h = Planck's constant and P = Linear momentum of a particle
- In terms of Energy, de Broglie wavelength:
$⇒ λ = \frac{h}{\sqrt{2mE}}$
Where m = mass and E = energy
EXPLANATION:
- In terms of Energy, de Broglie wavelength is given as,
$⇒ λ = \frac{h}{\sqrt{2mE}}$
$⇒ λ \propto \frac{1}{\sqrt{E}}$ -----(1)
- By equation 1 it is clear that the de Broglie wavelength is inversely proportional to the square root of the energy of the particle.
- Therefore if the energy of an electron becomes four times, then the de Broglie wavelength of the electron will become half. Hence, option 2 is correct.
Question 4
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Relation between wavelength of photon and electron of same energy is:

A
λ_ph > λ_e

B
λ_ph < λ_e

C
λ_ph = λ_e

D
$\frac {λ_e}{λ_{ph}} = constant$
Solution
CONCEPT:
- According to de Broglie matter has a dual nature of wave-particle.
- The wave associated with each moving particle is called matter waves.
- Characteristics of Matter waves:
  The lighter the particle, the greater is the de Broglie wavelength.
  The higher the velocity of the particle, the smaller is its de Broglie wavelength.
  The de Broglie wavelength of a particle is independent of the charge or nature of the particle.
  The matter waves are not electromagnetic in nature. Only charged particles produce electromagnetic waves.
- de Broglie wavelength associated with the particle
$⇒ λ = \frac{h}{P}$
Where, h = Planck's constant and P = Linear momentum of a particle
Photon
- Photon: Photon is an elementary particle that has a zero rest mass and moves with a speed of light in the vacuum.
  A photon is the “quantum of electromagnetic radiation”.
  It is the smallest and the fundamental particle of electromagnetic radiation.
CALCULATION:
- The wavelength of the electron,
$⇒ λ_{e}=\frac{h}{P_{e}}$ -----(1)
- The wavelength of the photon,
$⇒ λ_{Ph}=\frac{h}{P_{Ph}}$ -----(2)
- For the same energy, the momentum of the electron is more than that of the photon,
⇒ Pe > P_Ph -----(3)
From equation 1, equation 2, and equation 3,
⇒ λ_Ph> λ_e
- Hence, option 1 is correct.
Question 5
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If the intensity of incident photons is increased then the photoelectric current will-

A
Remain same

B
Decrease

C
Increase

D
First decrease and then increase
Solution
CONCEPT:
- Photoelectric effect: When a light of 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: The photocurrent may be stopped by applying a negative potential to anode w.r.t. cathode. The minimum potential required to stop the electron emitted from metal so that its kinetic energy becomes zero.
- Work function: It is the minimum amount of energy required so that metal emits an electron. It is represented with ϕ. Its unit is eV or joules.
  It is having different values for different metals.
Einstein’s equation for photoelectric effect is given by:
h ν = ϕ + K.E
Where h = planks constant = 6.6 × 10^-34 = 4.14 × 10^-15 eV-s, ν = incident frequency, ϕ = work function

EXPLANATION:
- Intensity (I) of photons is directly proportional to photoelectric current.
- When the intensity of incident photon increases, the number of electrons emitted from the metal surface increases due to which the photoelectric current increases. Hence option 3 is correct.
Question 6
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Two identical photo-electrode 1 and photo-electrode 2 receive the light of frequency f₁ and f₂ respectively of the same intensity. The photoelectric current in photo-electrode 1 and photo-electrode 2 is I₁ and I₂ respectively. If f₁ is greater than f₂, then:

A
I₁ = I₂

B
I1 > I2

C
I1 < I2

D
Can't predict
Solution
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.
- Einstein’s equation of photoelectric equation:
⇒ KEmax = hν - ϕo
Where h = Planck's constant, ν = frequency of incident radiation, ϕo = work function, and KE = maximum kinetic energy of electrons.
Laws of Photoelectric Effect:
1. For a light of any given frequency; (f > fTh) photoelectric current is directly proportional to the intensity of light.
2. For any given material, there is a certain minimum frequency, called threshold frequency, below which the emission of photoelectrons stops completely, no matter how high is the intensity of incident light.
3. The maximum kinetic energy of the photoelectrons is found to increase with the increase in the frequency of incident light, provided the frequency (f > fTh) exceeds the threshold limit. The maximum kinetic energy is independent of the intensity of light.
4. The photoelectric emission is an instantaneous process without any apparent time lag (10-9 sec or less), even when the intensity of the incident radiation is made extremely low.
EXPLANATION:
As we know that the photoelectric current is independent of the frequency of light.
For a light of any given frequency; (f > fTh) photoelectric current is directly proportional to the intensity of light.
∵ The intensity is the same for the light of frequency f₁ and f₂.
∴ I₁ = I₂
So the photoelectric current in both the electrodes will be equal. Hence, option 1 is correct.
Question 7
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According to the given statement:
a) Photoelectric effect can be explained by the wave theory of light
b) If the intensity of the incident light is increased the maximum kinetic energy of the photoelectrons also increases
Which of the following statement is correct?

A
Only a

B
Only b

C
Both a and b

D
Neither a nor b
Solution
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.
- Einstein’s equation of photoelectric effect:
⇒ KEmax = hν - ϕo
Where h = 6.63×10-34 J-sec = Planck's constant, ν = frequency of incident radiation, ϕo = work function, and KEmax = maximum kinetic energy of electrons.
EXPLANATION:
Photoelectric effect and wave theory of light:
- The phenomena of interference, diffraction, and polarisation were explained in a natural and satisfactory way by the wave picture of light.
  According to this picture, light is an electromagnetic wave consisting of electric and magnetic fields with a continuous distribution of energy over the region of space over which the wave is extended.
  According to the wave picture of light, the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously.
  The greater the intensity of radiation, the greater are the amplitude of electric and magnetic fields.
- Consequently, the greater the intensity, the greater should be the energy absorbed by each electron.
  In this picture, the maximum kinetic energy of the photoelectrons on the surface is then expected to increase with the increase in intensity.
- Also, no matter what the frequency of radiation is, a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons so that they exceed the minimum energy needed to escape from the metal surface.
- A threshold frequency, therefore, should not exist.
  But the experimental results show that the maximum kinetic energy of the photoelectrons does not depend on the intensity of radiation and the minimum frequency is required for the photoelectric effect.
- From the above explanation, we can say that the wave picture is unable to explain the most basic features of photoelectric emission.
- The experimental results show that the maximum kinetic energy of the photoelectrons does not depend on the intensity of radiation.
- So we can say that both the statements are wrong. Hence, option 4 is correct.
Question 8
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If electron, α particle, proton and neutron have the same kinetic energy, then the particle which has the shortest wavelength is:

A
Neutron

B
α Particle

C
Proton

D
Electron
Solution
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.
$λ = \frac{h}{\sqrt{2mE_k}}$
where λ is de Broglie wavelength, h is Planck's const, m is mass, E_k is the Kinetic energy.
- Mass of Proton = 1.673 × 10^-27 Kg;
- Mass of nutron = 1.675 × 10-27 Kg;
- Mass of α Particle = 4 × 1.673 × 10-27 Kg
- Mass of electron = 9.1 × 10-31 Kg
EXPLANATION:
- Given that the Kinetic energy of all the particles is the same. So
$λ = \frac{h}{\sqrt{2mE_k}} α \frac{1}{\sqrt m}$
- The above equation says the greater the mass the lesser the wavelength.
- Mass of α particle is greatest in the given option, So it will have the least wavelength.
- So the correct answer is option 2.
Question 9
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In the Davisson and Germer experiment, the velocity of electron emitted from the electron gun can be increased by

A
Decreasing the potential difference between the anode and filament

B
Increasing the potential difference between the anode and filament

C
Increasing the filament current

D
Decreasing the filament current
Solution
CONCEPT:
- Davison Germer experiment was conducted in 1927 to study the wave nature of the electron.
- The electrons are emitted by an electron gun and are accelerated by applying a suitable potential difference between anode and cathode.
- The finely collimated beam is directed towards the Ni crystal.
- The scattered electrons are collected using a movable detector and a graph is plotted between applied potential and angle of reflection.
- Analysis of those graphs showed the wave nature of the electron.
EXPLANATION:
- The velocity of an electron accelerated through a potential is given by
$V = √{\frac{2e V_o}{m}}$
Where e = Charge of electron and V_o = potential, m = mass of the particle
- From the above equation it is clear that
⇒ V ∝ √V_o
- The velocity of the electron is directly proportional to the square root of the potential difference between the anode and filament. Hence, option 2 is the answer
Question 10
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An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is

A
$\frac{1}{C}{\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$

B
${\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$

C
$c{\left( {2m\;E} \right)^{\frac{1}{2}}}$

D
$\frac{1}{C}{\left( {\frac{{2\;m}}{E}} \right)^{\frac{1}{2}}}$

Solution

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 (λ_e) is given by:
${\lambda _e} = \frac{h}{{\sqrt {2m\;E} }}$
Energy of a photon (E) = (hc)/λ
∴ The wavelength of the photon is
$\Rightarrow \lambda =\frac{{h\;c}}{E}$
Where E = energy, h = Planck constant, m = mass of electron and c = speed of light
Calculation:
Given - Energy of electron and photon is same (E)
The wavelength of the electron is
$\Rightarrow {\lambda _e} = \frac{h}{{\sqrt {2m\;E} }}$ ---------- (1)
The wavelength of the photon is
$\Rightarrow \lambda =\frac{{h\;c}}{E}$ ---------- (2)
To find the ratio of de-Broglie wavelength divides equation 1 by 2.
$\Rightarrow Ratio\;of\;de - Broglie\;wavelength = \frac{{Wavelenth\;of\;electron\;({\lambda _e})}}{{Wavelength\;of\;photon\;\left( \lambda \right)}} $
$\Rightarrow \lambda = \frac{{\frac{h}{{\sqrt {2m\;E} }}}}{{\frac{{h\;c}}{E}}} = \frac{1}{C}{\left( {\frac{E}{{2m}}} \right)^{\frac{1}{2}}}$