Dual Nature Of Radiation & Matter
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Which of the following represents the variation of de-Broglie wavelength (λ) with momentum (p) of a body?
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About This Quiz
This online test focuses on the Dual Nature of Radiation & Matter, assessing understanding of concepts like the de-Broglie wavelength, photoelectric effect, and quantum mechanics. It's designed for learners to evaluate their grasp on how matter and energy interact at quantum levels.
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- 2.
When an electron jumps across a potential difference of 1 V, it gains energy equal to:
Correct Answer
A.Explanation
When an electron jumps across a potential difference of 1 V, it gains energy equal to the electric potential difference. This is because the electric potential difference is a measure of the work done per unit charge to move the charge between two points in an electric field. Therefore, the energy gained by the electron is equal to the potential difference, which in this case is 1 V.Rate this question:
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- 3.
A graph plotted taking stopping potential along the y-axis and frequency of incident light (v) along the x-axis. The slope of this graph is given by:
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H
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E
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Eh
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H/e
Correct Answer
A. H/eExplanation
The slope of the graph represents the ratio of Planck's constant (h) to the charge of an electron (e). This is because the stopping potential is directly proportional to the frequency of incident light, which is related to the energy of the photons. According to the photoelectric effect equation, the energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency. The stopping potential is directly proportional to the energy of the photons, so the slope of the graph gives the ratio of h to e.Rate this question:
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- 4.
Photoelectric effect can be caused by:
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X-rays
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Visible light
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γ-rays
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All of these
Correct Answer
A. All of theseExplanation
The photoelectric effect is the emission of electrons from a material when it absorbs electromagnetic radiation. It can be caused by X-rays, visible light, and gamma rays, as all of these forms of radiation have enough energy to dislodge electrons from their atoms. This phenomenon is not limited to a specific type of radiation but can occur with any form of electromagnetic radiation that has sufficient energy.Rate this question:
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- 5.
Which of the following has minimum stopping potential?
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Blue
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Yellow
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Violet
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Red
Correct Answer
A. RedExplanation
The question is asking for the color that has the minimum stopping potential. Stopping potential refers to the minimum potential difference required to stop the flow of electrons in a photoelectric effect experiment. According to the photoelectric effect, the stopping potential is directly proportional to the frequency of the incident light. The frequency of light is inversely proportional to its wavelength. Since red light has the longest wavelength among the given options, it has the lowest frequency and therefore requires the minimum stopping potential to stop the flow of electrons.Rate this question:
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- 6.
The kinetic energy of an electron gets tripled, then the de-Broglie wavelength associated with it changes by a factor:
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![Dual Nature Of Radiation & Matter - Quiz]()
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![Dual Nature Of Radiation & Matter - Quiz]()
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![Dual Nature Of Radiation & Matter - Quiz]()
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3
Correct Answer
A.Explanation
When the kinetic energy of an electron triples, its velocity also triples since kinetic energy is directly proportional to the square of velocity. The de-Broglie wavelength of an electron is inversely proportional to its velocity. Therefore, when the velocity triples, the de-Broglie wavelength changes by a factor of 1/3.Rate this question:
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- 7.
Maximum kinetic energy (Eo) of a photoelectron varies with frequency (f) of the incident radiation. Which of the following correctly represents this variation?
Correct Answer
A.Explanation
The maximum kinetic energy of a photoelectron is directly proportional to the frequency of the incident radiation. This means that as the frequency increases, the maximum kinetic energy of the photoelectron also increases.Rate this question:
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- 8.
When the radiation of a photoelectron emitter, the stopping potential is found to be 9V. If e/m for the electron is the maximum velocity of the ejected electron is:
Correct Answer
A.Explanation
The stopping potential in a photoelectric experiment is the minimum potential required to stop the photoelectrons from reaching the collector plate. It is given by the equation V = eV0, where V is the stopping potential and e is the charge of an electron. Since the stopping potential is given as 9V, we can rearrange the equation to solve for V0. The maximum velocity of the ejected electron can be determined using the equation V0 = √(2eV0/m), where m is the mass of the electron. Given that e/m is missing from the question, it is not possible to calculate the maximum velocity of the ejected electron.Rate this question:
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- 9.
Protons and alpha particles have the same de-Broglie wavelength. What is same for both of them?
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Energy
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Time Period
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Frequency
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Momentum
Correct Answer
A. MomentumExplanation
The de-Broglie wavelength is inversely proportional to the momentum of a particle. Since protons and alpha particles have the same de-Broglie wavelength, it implies that they must have the same momentum. Therefore, momentum is the same for both protons and alpha particles.Rate this question:
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- 10.
Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is:
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1 : 4
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1 : 2
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1 : 1
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1 : 5
Correct Answer
A. 1 : 2Explanation
The ratio of the maximum speeds of the emitted electrons is 1:2. This can be explained using the photoelectric effect. The work function of the metallic surface is the minimum energy required to remove an electron from the surface. When a photon with energy equal to or greater than the work function strikes the surface, an electron is emitted with a maximum kinetic energy equal to the difference between the photon energy and the work function. In this case, the first radiation with energy 1 eV is less than the work function, so no electrons are emitted. However, the second radiation with energy 2.5 eV is greater than the work function, so electrons are emitted with a maximum kinetic energy of 2.5 eV - 0.5 eV = 2 eV. Therefore, the ratio of the maximum speeds of the emitted electrons is 1:2.Rate this question:
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- 11.
A proton, a neutron, an electron and an α-particle have same energy. Then their de-Broglie wavelengths compare as:
Correct Answer
A.Explanation
The de-Broglie wavelength is inversely proportional to the momentum of a particle. Since all the particles mentioned (proton, neutron, electron, and α-particle) have the same energy, their momenta will be different due to their different masses. The momentum of a particle is given by its mass times its velocity, so the lighter the particle, the higher its velocity and momentum. Therefore, the electron, being the lightest particle, will have the highest momentum and the shortest de-Broglie wavelength, while the α-particle, being the heaviest, will have the lowest momentum and the longest de-Broglie wavelength.Rate this question:
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- 12.
An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Ǻ. What is the maximum energy of a photon in the radiation?
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![Dual Nature Of Radiation & Matter - Quiz]()
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![Dual Nature Of Radiation & Matter - Quiz]()
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![Dual Nature Of Radiation & Matter - Quiz]()
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None of these
Correct Answer
A.Explanation
The maximum energy of a photon in the radiation can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Since the short wavelength end is given as 0.45 Ǻ, we can substitute this value into the equation to find the maximum energy of the photon.Rate this question:
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- 13.
If the momentum of electron is changed by P, then the de Broglie wavelength associated with it changes by 0.5%. The initial momentum of electron will be:
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100 P
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200 P
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400 P
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P/200
Correct Answer
A. 200 PExplanation
The de Broglie wavelength of a particle is inversely proportional to its momentum. Therefore, if the de Broglie wavelength changes by 0.5% when the momentum changes by P, it means that the momentum has doubled. Since the initial momentum is P, the correct answer is 200 P.Rate this question:
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- 14.
The de-Broglie wavelength of particle of mass 1 mg moving with a velocity of 1 ms-1, in terms of Planck’s constant h, is given by (in metre):
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105 h
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106 h
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10-3 h
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103 h
Correct Answer
A. 106 hExplanation
The de-Broglie wavelength of a particle is given by the equation λ = h/mv, where λ is the wavelength, h is Planck's constant, m is the mass of the particle, and v is its velocity. In this case, the mass of the particle is 1 mg (0.001 g) and its velocity is 1 m/s. Plugging these values into the equation, we get λ = h/(0.001 g * 1 m/s) = h/(0.001 kg m/s) = 1000h/(kg m/s) = 1000h. Therefore, the correct answer is 1000h, which is equivalent to 106h.Rate this question:
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- 15.
What is the de-Broglie wavelength of a proton accelerated from rest through a potential difference of V volts?
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12.3 Ǻ
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1.23 Ǻ
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0.123 Ǻ
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None of these
Correct Answer
A. 1.23 ǺExplanation
The de-Broglie wavelength of a particle is given by the equation λ = h/p, where λ is the wavelength, h is the Planck's constant, and p is the momentum of the particle. In this case, the proton is accelerated through a potential difference, which implies that it gains kinetic energy. The momentum of the proton can be calculated using the equation p = √(2mE), where m is the mass of the proton and E is the kinetic energy gained. Since the proton is initially at rest, the kinetic energy gained is equal to the potential energy gained, which is given by E = qV, where q is the charge of the proton and V is the potential difference. Substituting these values into the equation for momentum and then into the equation for wavelength, we find that the de-Broglie wavelength of the proton is 1.23 Ǻ.Rate this question:
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Mar 21, 2023Quiz Edited by
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Jan 20, 2014Quiz Created by
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