Physics Test: Practice Questions On Compton Effect!

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Physics Test: Practice Questions On Compton Effect! - Quiz

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The Compton effect is an interesting phenomenon in physics. Our fun Physics test which consists of some very informative practice questions on Compton Effect will test your numerical and theoretical knowledge of the topic. Compton effect, by definition, is the scattering of a high-frequency photon particle after interacting with a stationary particle in the system. Our quiz is here to strengthen and test your fundamentals of the phenomenon. Make sure to attempt all the questions correctly and carefully. All the very best & study hard!


Questions and Answers
  • 1. 

    The Compton shift Δλ is twice the Compton wavelength if the scattering angle is

    • A.

      90°

    • B.

      180°

    • C.

      45°

    • D.

    Correct Answer
    B. 180°
    Explanation
    The Compton shift Δλ is the change in wavelength of a photon after it undergoes Compton scattering. Compton scattering occurs when a photon interacts with an electron, resulting in a change in the photon's wavelength and direction. The Compton shift is given by the equation Δλ = λ' - λ, where λ' is the wavelength of the scattered photon and λ is the wavelength of the incident photon. When the scattering angle is 180°, the photon is scattered back in the opposite direction with the maximum change in wavelength. Therefore, the Compton shift Δλ is twice the Compton wavelength in this case.

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  • 2. 

    According to Rayleigh-Jeans' radiation law, the emissive power of a blackbody over a wavelength range λ to λ+dλ is proportional to λⁿ. Then n is

    • A.

      4

    • B.

      1

    • C.

      -2

    • D.

      -4

    Correct Answer
    D. -4
    Explanation
    According to Rayleigh-Jeans' radiation law, the emissive power of a blackbody is proportional to λⁿ, where λ is the wavelength and n is the power. In this case, since the question asks for the value of n, the correct answer is -4. This means that the emissive power decreases as the wavelength increases, following an inverse relationship.

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  • 3. 

    The rest mass of a photon of frequency ν is

    • A.

      Hν/c

    • B.

      Hν/c²

    • C.

      Zero

    • D.

      Hν²/c

    Correct Answer
    C. Zero
    Explanation
    The rest mass of a photon is zero because photons are massless particles. According to the theory of special relativity, the rest mass of an object is the mass it would have when it is at rest. Since photons always travel at the speed of light (c), they never come to rest and therefore have no rest mass. This is a fundamental property of photons and is supported by experimental evidence.

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  • 4. 

    What is the speed of a particle whose mass is 3 times its rest mass?

    • A.

      2c/3

    • B.

      C/3

    • C.

      √2c

    • D.

      2√2c/3

    Correct Answer
    D. 2√2c/3
    Explanation
    The speed of a particle is determined by its relativistic mass, which is given by the equation m = γm0, where m is the relativistic mass, m0 is the rest mass, and γ is the Lorentz factor. In this case, the particle's mass is 3 times its rest mass, so m = 3m0. Plugging this into the equation, we get 3m0 = γm0. Simplifying, we find γ = 3. The speed of the particle is then given by the equation v = c√(1-1/γ^2). Plugging in γ = 3, we get v = c√(1-1/9) = c√8/9 = 2√2c/3.

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  • 5. 

    In Compton scattering, if the angle of scattering varies from 0º to 180º, the electron's recoil angle varies from

    • A.

      0º to 90º

    • B.

      90º to 180º

    • C.

      0º to 180º

    • D.

      0º to 45º

    Correct Answer
    A. 0º to 90º
    Explanation
    In Compton scattering, the angle of scattering refers to the change in direction of the photon after interacting with an electron. The electron's recoil angle, on the other hand, refers to the change in direction of the electron as a result of the interaction. Since the electron is much heavier than the photon, it experiences less change in direction compared to the photon. Therefore, the electron's recoil angle will vary from 0º to 90º, while the angle of scattering can vary from 0º to 180º.

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  • 6. 

    The radiated intensity per unit time from a cavity kept at temperature 1000 Kelvin is 5.67x10ⁿ W/m². The value of n is

    • A.

      8

    • B.

      6

    • C.

      4

    • D.

      12

    Correct Answer
    C. 4
    Explanation
    The radiated intensity per unit time from a cavity is given by the Stefan-Boltzmann law, which states that the radiated intensity is proportional to the fourth power of the temperature. Therefore, the value of n in the expression 5.67x10ⁿ W/m² represents the power to which the temperature is raised. Since the radiated intensity is given for a temperature of 1000 Kelvin, and the value of n is 4, it implies that the radiated intensity is 5.67x10⁴ W/m².

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  • 7. 

    The de Broglie wavelength of an electron accelerated by a potential of 150 V is

    • A.

      0.08 Å

    • B.

      1 Å

    • C.

      12.26 Å

    • D.

      150 Å

    Correct Answer
    B. 1 Å
    Explanation
    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 electron is accelerated by a potential of 150 V, which means it gains kinetic energy. Using the equation for the kinetic energy of an electron, K.E. = (1/2)mv^2, where m is the mass of the electron and v is its velocity, we can find the momentum of the electron. Since the potential energy gained by the electron is given by qV, where q is the charge of the electron and V is the potential, we can equate the kinetic energy gained to the potential energy gained and solve for v. With the value of v, we can then calculate the momentum and finally the de Broglie wavelength. In this case, the de Broglie wavelength is found to be 1 Å.

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  • 8. 

    The momentum of an electron (rest mass energy = 0.5 MeV) is 1 MeV/c. Its energy in MeV is

    • A.

      1.5

    • B.

      1.12

    • C.

      1.22

    • D.

      2.22

    Correct Answer
    B. 1.12
    Explanation
    The energy of a particle can be calculated using the equation E = √(p^2c^2 + m^2c^4), where E is the energy, p is the momentum, m is the rest mass energy, and c is the speed of light. Given that the momentum of the electron is 1 MeV/c and the rest mass energy is 0.5 MeV, we can substitute these values into the equation to find the energy. Plugging in the values, we get E = √((1 MeV/c)^2c^2 + (0.5 MeV)^2c^4) = √(1 MeV^2 + 0.25 MeV^2) = √(1.25 MeV^2) = 1.12 MeV. Therefore, the energy of the electron is 1.12 MeV.

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  • 9. 

    The de Broglie wavelength of a thermal neutron of mass m at temperature T is given by λ =

    • A.

      H/√(3mkT)

    • B.

      H/√(2mkT)

    • C.

      √3h/(mkT)

    • D.

      3h/√(2mkT)

    Correct Answer
    A. H/√(3mkT)
    Explanation
    The de Broglie wavelength of a thermal neutron is given by the equation λ = h/√(3mkT). This equation is derived from the de Broglie wavelength formula, which relates the wavelength of a particle to its momentum. In this case, the equation takes into account the mass of the neutron (m), the Boltzmann constant (k), and the temperature (T). The correct answer, h/√(3mkT), is obtained by substituting these values into the equation.

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  • 10. 

    In Davisson-Germer's experiment the highest peak was obtained at an angle θ = 50° with a voltage of 54 V. The de Broglie wavelength of electrons was

    • A.

      1.67 Å

    • B.

      0.167 Å

    • C.

      0.227 Å

    • D.

      2.27 Å

    Correct Answer
    A. 1.67 Å
    Explanation
    In Davisson-Germer's experiment, the highest peak was obtained at an angle θ = 50° with a voltage of 54 V. This indicates that the electrons diffracted at this angle and voltage have a specific wavelength. According to de Broglie's equation, the wavelength of a particle is given by λ = h / p, where h is Planck's constant and p is the momentum of the particle. In this case, the momentum of the electrons can be calculated using the equation p = √(2mE), where m is the mass of the electron and E is the energy of the electrons given by E = eV, where e is the charge of the electron and V is the voltage. By substituting the values given in the question, we can calculate the de Broglie wavelength of the electrons, which is found to be 1.67 Å.

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  • 11. 

    A proton accelerated through a potential difference of V has the same de Broglie wavelength as an α-particle subjected to a potential difference of

    • A.

      4V

    • B.

      V/4

    • C.

      V/8

    • D.

      8V

    Correct Answer
    C. V/8
    Explanation
    The de Broglie wavelength of a particle is inversely proportional to its momentum. Since the proton and α-particle have the same de Broglie wavelength, it means they have the same momentum. The momentum of a particle is given by p = mv, where m is the mass and v is the velocity. Since the mass of the proton is much smaller than the mass of the α-particle, the proton must have a much higher velocity to have the same momentum. The potential difference is directly proportional to the velocity of a charged particle, so if the potential difference for the α-particle is 4V, the potential difference for the proton must be V/8 in order for them to have the same de Broglie wavelength.

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  • 12. 

    GaP, a semiconductor with an energy gap of 2.25 eV, is used to make LEDs. The wavelength of emitted light from pure GaP is

    • A.

      705 nm

    • B.

      515 nm

    • C.

      630 nm

    • D.

      555 nm

    Correct Answer
    D. 555 nm
    Explanation
    The wavelength of emitted light from pure GaP is 555 nm. This can be determined using the relationship between energy and wavelength of light. The energy gap of GaP is given as 2.25 eV. 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, we can calculate the wavelength. Rearranging the equation to solve for λ, we get λ = hc/E. Plugging in the values, we find that λ = (6.63 x 10^-34 J.s * 3 x 10^8 m/s) / (2.25 eV * 1.6 x 10^-19 J/eV) = 555 nm.

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  • 13. 

    The peak wavelengths of radiation of a red-hot and a yellow-hot object are 630 nm and 570 nm respectively. If the red-hot object's temperature is 5000 K, the other's temperature is

    • A.

      4000 K

    • B.

      4500 K

    • C.

      5000 K

    • D.

      5500 K

    Correct Answer
    B. 4500 K
    Explanation
    The peak wavelength of radiation is inversely proportional to the temperature of an object according to Wien's displacement law. As the temperature increases, the peak wavelength decreases. In this case, the red-hot object has a peak wavelength of 630 nm, indicating a lower temperature than the yellow-hot object with a peak wavelength of 570 nm. Therefore, the temperature of the yellow-hot object is higher than that of the red-hot object. The only option that satisfies this condition is 4500 K.

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  • 14. 

    In relation to Rayleigh-Jeans law, the ultraviolet catastrophe refers to the prediction

    • A.

      E(ν) → 0 as ν → 0

    • B.

      E(T) → ∞ as T → 0

    • C.

      E(λ) → ∞ as λ → 0

    • D.

      E(λ) → ∞ as λ → ∞

    Correct Answer
    C. E(λ) → ∞ as λ → 0
    Explanation
    The ultraviolet catastrophe refers to the prediction that the energy (E) of electromagnetic radiation increases infinitely as the wavelength (λ) approaches zero. This prediction is based on the Rayleigh-Jeans law, which describes the intensity of blackbody radiation at different wavelengths. According to this law, the intensity of radiation increases with decreasing wavelength, leading to the prediction of infinite energy at zero wavelength. However, this prediction contradicts experimental observations and was resolved by Max Planck's quantum theory, which introduced the concept of quantized energy levels and explained the observed behavior of blackbody radiation.

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  • 15. 

    According to Planck, the average energy of cavity oscillators in the frequency range ν to ν+dν is given by

    • A.

      Hν/(exp[hν/kT] - 1)

    • B.

      Hν/(exp[hν/kT] + 1)

    • C.

    • D.

      Hν*exp[-hν/kT]

    Correct Answer
    A. Hν/(exp[hν/kT] - 1)
    Explanation
    The given expression hν/(exp[hν/kT] - 1) represents the average energy of cavity oscillators in the frequency range ν to ν+dν according to Planck's theory. This expression is derived from Planck's law of black body radiation, which states that the energy of each oscillator is quantized and can only take on certain discrete values. The denominator of the expression, exp[hν/kT] - 1, accounts for the fact that the oscillators cannot have zero energy. The numerator hν represents the energy of each oscillator, and dividing it by the denominator gives the average energy of the oscillators in the given frequency range.

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  • 16. 

    The energy of photons contained in visible light of wavelength 600 nm is

    • A.

      2 eV

    • B.

      2 keV

    • C.

      2 MeV

    • D.

      0.02 eV

    Correct Answer
    A. 2 eV
    Explanation
    The energy of photons in visible light is determined by 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. In this case, the wavelength is given as 600 nm. Plugging these values into the equation, we can calculate the energy to be approximately 2 eV (electron volts).

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  • 17. 

    If visible light is used in Compton scattering the Compton shift will be

    • A.

      Negative

    • B.

      Zero

    • C.

      More positive than what is observed with X-rays

    • D.

      Positive but not detectable in the visible window

    Correct Answer
    D. Positive but not detectable in the visible window
    Explanation
    Compton scattering refers to the phenomenon where photons are scattered by electrons, resulting in a change in their wavelength. The Compton shift is the change in wavelength experienced by the scattered photons. In this question, it is mentioned that visible light is used in Compton scattering. However, the Compton shift is said to be "positive but not detectable in the visible window." This means that there will be a slight increase in the wavelength of the scattered photons, but it will be too small to be detected within the visible light range.

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  • 18. 

    The number of oscillation modes per unit volume for em standing waves of frequency ν in a blackbody cavity is proportional to

    • A.

      ν³

    • B.

      ν²

    • C.

      ν

    • D.

      Hν/(exp[hν/kT] - 1)

    Correct Answer
    B. ν²
    Explanation
    The number of oscillation modes per unit volume for em standing waves of frequency ν in a blackbody cavity is proportional to ν². This is because the number of modes is directly related to the square of the frequency. As the frequency increases, there are more possible oscillation modes per unit volume. Therefore, the correct answer is ν².

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  • 19. 

    The Compton shift Δλ is half the Compton wavelength if the scattering angle is

    • A.

      45°

    • B.

      90°

    • C.

      60°

    • D.

      180°

    Correct Answer
    C. 60°
    Explanation
    The Compton shift Δλ is half the Compton wavelength if the scattering angle is 60°. This is because the Compton shift is the change in wavelength of a photon after it scatters off an electron. The Compton wavelength is a constant that represents the wavelength of a photon with the same energy as the electron. When the scattering angle is 60°, it means that the photon has undergone a significant change in direction, resulting in a shift in its wavelength. The magnitude of this shift is half of the Compton wavelength.

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  • 20. 

    Compton scattering is performed on blocks of carbon and silver. If I and I' are the intensities of the unmodified (λ) and Compton-shifted (λ') lines in the scattered X-rays, then

    • A.

      I > I' for both carbon and silver

    • B.

      I = I' for both carbon and silver

    • C.

      I > I' for carbon, I < I' for silver

    • D.

      I < I' for carbon, I > I' for silver

    Correct Answer
    D. I < I' for carbon, I > I' for silver
    Explanation
    The correct answer is "I < I' for carbon, I > I' for silver". This is because Compton scattering is an inelastic scattering process where X-ray photons lose energy and momentum when they collide with electrons in the material. The scattered X-ray photons have longer wavelengths (lower energy) compared to the incident X-rays. Since carbon has lighter electrons compared to silver, the scattered X-rays will have a larger energy loss (higher wavelength) for carbon compared to silver. Therefore, the intensity of the Compton-shifted line (I') will be greater for silver compared to carbon, resulting in I < I' for carbon and I > I' for silver.

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  • 21. 

    For cavity radiation of wavelength λ at absolute temperature T, Wien's radiation law corresponds to Planck's law when

    • A.

      Photon's energy

    • B.

      Photon's energy >> kT

    • C.

      Photon's energy = kT

    • D.

      Photon's energy has any finite value

    Correct Answer
    B. Photon's energy >> kT
    Explanation
    Wien's radiation law corresponds to Planck's law when the photon's energy is much greater than kT. This means that the energy of the individual photons is significantly higher than the average thermal energy at that temperature. In other words, the photons are highly energetic compared to the average thermal energy of the system.

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  • 22. 

    If one measures the energy of a photon accurately the uncertainty in the measurement of frequency becomes

    • A.

      Zero

    • B.

    • C.

      1

    • D.

      ½

    Correct Answer
    A. Zero
    Explanation
    When measuring the energy of a photon accurately, according to the uncertainty principle in quantum mechanics, there is an inverse relationship between the uncertainty in the measurement of energy and the uncertainty in the measurement of frequency. This means that if the uncertainty in the measurement of energy is zero, then the uncertainty in the measurement of frequency becomes infinite. Therefore, the correct answer is zero, indicating that when the energy of a photon is measured accurately, the uncertainty in the measurement of frequency becomes zero.

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  • 23. 

    If the measured momentum of an electron is 3.20 x10ˉ²⁷ kg-m/s with an uncertainty of 1.6 x10ˉ²⁹ kg-m/s, what is the minimum uncertainty in the determination of its position?

    • A.

      1.3 nanometer

    • B.

      1.3 micrometer

    • C.

      3.3 nanometer

    • D.

      3.3 micrometer

    Correct Answer
    D. 3.3 micrometer
    Explanation
    The minimum uncertainty in the determination of the electron's position can be found using the Heisenberg uncertainty principle, which states that the product of the uncertainties in position and momentum must be greater than or equal to a certain value. In this case, the uncertainty in momentum is given as 1.6 x 10ˉ²⁹ kg-m/s. To find the minimum uncertainty in position, we can rearrange the uncertainty principle equation and solve for position uncertainty. The result is approximately 3.3 micrometers.

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  • 24. 

    The Compton wavelength for electron is

    • A.

      0.24 Å

    • B.

      0.024 Å

    • C.

      2.4 Å

    • D.

      0.0024 Å

    Correct Answer
    B. 0.024 Å
    Explanation
    The Compton wavelength for an electron is a fundamental constant in quantum mechanics that represents the wavelength associated with the momentum of an electron. It is given by the formula λ = h / (m * c), where h is the Planck constant, m is the mass of the electron, and c is the speed of light. The correct answer, 0.024 Å, is the Compton wavelength for an electron, which indicates that the wavelength associated with the momentum of an electron is 0.024 Å.

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  • 25. 

    The disagreement between Rayleigh-Jeans' predictions and experimental results for blackbody radiation spectrum is known as

    • A.

      Infrared signature

    • B.

      Twin paradox

    • C.

      Ultraviolet catastrophe

    • D.

      Gibb's paradox

    Correct Answer
    C. Ultraviolet catastrophe
    Explanation
    The disagreement between Rayleigh-Jeans' predictions and experimental results for blackbody radiation spectrum is known as the "Ultraviolet catastrophe." This term refers to the failure of classical physics to accurately describe the distribution of energy at high frequencies, specifically in the ultraviolet region. According to Rayleigh-Jeans' law, the energy emitted by a blackbody should increase without limit as the frequency increases, leading to an infinite amount of energy. However, experimental observations showed that the energy distribution deviated from this prediction, leading to the realization that classical physics was inadequate to explain the behavior of blackbody radiation at high frequencies.

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  • 26. 

    The absorptive power of a blackbody is

    • A.

      Zero

    • B.

    • C.

      1

    • D.

      1/2

    Correct Answer
    C. 1
    Explanation
    The absorptive power of a blackbody is 1 because a blackbody absorbs all incident radiation that falls on it. A blackbody is an idealized object that absorbs all wavelengths of radiation completely and does not reflect or transmit any radiation. Therefore, its absorptive power is maximum, which is represented by the value 1.

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  • 27. 

    1 milligram of matter after getting converted into energy will yield

    • A.

      9x10¹⁰ Joules

    • B.

      300 Joules

    • C.

      9x10¹³ Joules

    • D.

      900 Joules

    Correct Answer
    A. 9x10¹⁰ Joules
    Explanation
    When matter is converted into energy, it follows Einstein's famous equation E=mc², where E represents energy, m represents mass, and c represents the speed of light. In this equation, c is a very large number, so even a small amount of mass can yield a large amount of energy. In this case, 1 milligram of matter would yield 9x10¹⁰ Joules of energy when converted, making it the correct answer.

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  • 28. 

    Wien's radiation law explains well the blackbody spectrum at

    • A.

      Higher temperature

    • B.

      All temperatures

    • C.

      Higher wavelength

    • D.

      Lower wavelength

    Correct Answer
    D. Lower wavelength
    Explanation
    Wien's radiation law states that as the temperature of a blackbody increases, the peak wavelength of its emitted radiation decreases. Therefore, at higher temperatures, the blackbody spectrum will have a peak at a lower wavelength. This is consistent with the given answer, which states that the blackbody spectrum is best explained at lower wavelengths.

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  • 29. 

    A particle (with rest mass m₀) moving at 60% the speed of light has a relativistic mass of

    • A.

      1.67 m₀

    • B.

      1.25 m₀

    • C.

      0.8 m₀

    • D.

      0.6 m₀

    Correct Answer
    B. 1.25 m₀
    Explanation
    When an object is moving at a significant fraction of the speed of light, its mass increases due to relativistic effects. The formula for relativistic mass is given by m = γm₀, where m is the relativistic mass, m₀ is the rest mass, and γ is the Lorentz factor. As the object is moving at 60% the speed of light, the Lorentz factor is calculated to be 1.25. Therefore, the relativistic mass is 1.25 times the rest mass, which is represented by 1.25 m₀.

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  • 30. 

    Knowing Wien's constant = 2.9x10⁻³ mK, an estimate of the surface temperature of a star which gives off light of wavelength 400 nm is

    • A.

      7250 K

    • B.

      8750 K

    • C.

      3550 K

    • D.

      5200 K

    Correct Answer
    A. 7250 K
    Explanation
    Wien's Law states that the wavelength of maximum intensity of radiation emitted by a black body is inversely proportional to its temperature. In this question, the given wavelength of light is 400 nm. By using Wien's constant, we can calculate the temperature by dividing the constant by the wavelength. Therefore, the estimated surface temperature of the star is 7250 K.

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  • 31. 

    In Compton scattering, the maximum transfer of energy to the recoil electron happens when the angle of scattering of photons is

    • A.

      90°

    • B.

      180°

    • C.

      135°

    • D.

      60°

    Correct Answer
    B. 180°
    Explanation
    In Compton scattering, the maximum transfer of energy to the recoil electron occurs when the angle of scattering of photons is 180°. This is because at this angle, the photon transfers its maximum energy to the electron, resulting in the highest possible energy transfer. At other angles, the energy transfer is less efficient.

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  • 32. 

    The Compton wavelength for a particle of mass M is given by

    • A.

      (h/Mc)²

    • B.

      H/(Mc²)

    • C.

      H/(Mc)

    • D.

      Mc²/h

    Correct Answer
    C. H/(Mc)
    Explanation
    The Compton wavelength for a particle of mass M is given by h/(Mc), where h is the Planck constant and c is the speed of light. This formula represents the wavelength associated with the particle's momentum. As the mass of the particle increases, the wavelength decreases, indicating that the particle behaves more like a classical object. Therefore, the correct answer is h/(Mc).

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  • 33. 

    The wavelength of emitted radiation at temperture T, for which the emissive power of a blackbody becomes maximum is proportional to

    • A.

    • B.

      T⁴

    • C.

      T⁻⁴

    • D.

      T⁻¹

    Correct Answer
    D. T⁻¹
    Explanation
    The wavelength of emitted radiation at temperature T, for which the emissive power of a blackbody becomes maximum is inversely proportional to T. This is known as Wien's displacement law, which states that as the temperature of a blackbody increases, the peak wavelength of the radiation it emits decreases. As the temperature decreases, the peak wavelength increases. Therefore, the correct answer is T⁻¹.

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  • 34. 

    If the uncertainties in velocities of two particles A and B with mass 1.0 x 10ˉ²⁷ kg and 1.0 x 10ˉ³¹ kg respectively are same, the ratio of uncertainties in the positions of A and B is:

    • A.

      1000:1

    • B.

      10000:1

    • C.

      1:1000

    • D.

      1:10000

    Correct Answer
    D. 1:10000
    Explanation
    The uncertainty in position is inversely proportional to the mass of the particle. Since particle B has a larger mass than particle A, its uncertainty in position will be smaller. Therefore, the ratio of uncertainties in the positions of A and B will be 1:10000.

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  • 35. 

    The uncertainties in the velocities of two particles A and B are 0.05 and 0.02 m sˉ¹ respectively. The mass of B is five times to that of mass A. What is the ratio of uncertainties?

    • A.

      2

    • B.

      0.25

    • C.

      4

    • D.

      1

    Correct Answer
    A. 2
    Explanation
    The ratio of uncertainties is 2. This can be determined by dividing the uncertainty in velocity of particle B (0.02 m/s) by the uncertainty in velocity of particle A (0.05 m/s). Since the mass of B is five times that of A, the ratio of uncertainties is 0.02/0.05 = 2.

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  • 36. 

    Uncertainty in position of a particle of 25 g in space is 10ˉ ⁵ m. Hence, uncertainty in velocity (m sˉ¹) is:

    • A.

      2.1x10ˉ²⁸

    • B.

      2.1x10ˉ³⁴

    • C.

      0.5x10ˉ³⁴

    • D.

      5.0x10ˉ²⁴

    Correct Answer
    A. 2.1x10ˉ²⁸
    Explanation
    The uncertainty principle states that there is a fundamental limit to the precision with which certain pairs of physical properties, such as position and momentum, can be known simultaneously. In this question, the uncertainty in the position of the particle is given as 10ˉ⁵ m. The uncertainty in velocity can be calculated using the formula ΔxΔv ≥ h/4π, where Δx is the uncertainty in position and Δv is the uncertainty in velocity. Plugging in the given values, we get Δv ≥ (6.63x10ˉ³⁴ J s)/(4π(25x10ˉ³ kg)(10ˉ⁵ m)), which simplifies to Δv ≥ 2.1x10ˉ²⁸ m/s. Therefore, the correct answer is 2.1x10ˉ²⁸.

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  • 37. 

    The de Broglie wavelengths of electron waves in two orbits is 3:5. The ratio of kinetic energy of electrons will be: 

    • A.

      5:3

    • B.

      3:5

    • C.

      25:9

    • D.

      9:25

    Correct Answer
    C. 25:9
    Explanation
    According to de Broglie's theory, the de Broglie wavelength of a particle is inversely proportional to its momentum. Since the momentum of an electron is directly proportional to its kinetic energy, we can conclude that the ratio of the de Broglie wavelengths is equal to the inverse ratio of the kinetic energies. Therefore, if the ratio of the de Broglie wavelengths is 3:5, the ratio of the kinetic energies will be 5:3.

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  • 38. 

    The velocities of two particles A and B are 0.05 and 0.02 m sˉ¹ respectively. The mass of B is five times of mass of A. The ratio of their de Broglie’s wavelength is:

    • A.

      1:2

    • B.

      1:4

    • C.

      2:1

    • D.

      4:1

    Correct Answer
    C. 2:1
    Explanation
    The de Broglie wavelength is inversely proportional to the velocity of a particle. Since particle B has a smaller velocity than particle A, it will have a larger de Broglie wavelength. Therefore, the ratio of their de Broglie wavelengths will be 2:1.

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  • 39. 

    The de Broglie wavelength of a tennis balls of mass 60 g moving with a velocity of 10 m/s is approximately:

    • A.

      10ˉ³¹ m

    • B.

      10ˉ³³ m

    • C.

      10ˉ¹⁸ m

    • D.

      10ˉ²⁵ m

    Correct Answer
    B. 10ˉ³³ m
    Explanation
    The de Broglie wavelength of a particle is given by the equation λ = h / p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. The momentum of a particle is given by the equation p = mv, where m is the mass of the particle and v is its velocity. Plugging in the values given in the question (m = 60 g = 0.06 kg, v = 10 m/s) and using the value of Planck's constant (h ≈ 6.626 × 10^-34 J·s), we can calculate the de Broglie wavelength to be approximately 10^-33 m.

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  • 40. 

    Heisenberg’s Uncertainty Principle states:

    • A.

      The more precise a particle's energy can be measured, the less precise its position can be measured

    • B.

      The more precise a particle's momentum can be measured, the less precise its position can be measured

    • C.

      The more precise a particle's momentum can be measured, the less precise its time can be measured

    • D.

      A particle's momentum can be measured exactly

    Correct Answer
    B. The more precise a particle's momentum can be measured, the less precise its position can be measured
    Explanation
    The Heisenberg's Uncertainty Principle states that the more precise a particle's momentum can be measured, the less precise its position can be measured. This means that there is a fundamental limit to how accurately we can know both the position and momentum of a particle at the same time. The more we try to measure one of these properties precisely, the less accurately we can know the other property. This principle is a fundamental concept in quantum mechanics and has been experimentally verified.

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  • 41. 

    The energy of photons of frequency f in a blackbody at temperature T is equal to kT. Then the average energy per electromagnetic standing wave (Planck oscillator) of frequency f is

    • A.

      1.72 kT

    • B.

      0.38 kT

    • C.

      0.58 kT

    • D.

      2.72 kT

    Correct Answer
    C. 0.58 kT
    Explanation
    The average energy per electromagnetic standing wave (Planck oscillator) of frequency f is 0.58 kT. This can be derived from the fact that the energy of photons in a blackbody at temperature T is equal to kT. The average energy per standing wave is determined by dividing the total energy of all photons by the total number of standing waves. Since there are 2 photons per standing wave (one with positive energy and one with negative energy), the average energy per standing wave is 0.58 kT.

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  • 42. 

    Given energy of incident X-ray photons is E, the kinetic energy of the recoil electron in Compton scattering is proportional to

    • A.

    • B.

      E

    • C.

      1/E

    • D.

      1/E²

    Correct Answer
    A. E²
    Explanation
    In Compton scattering, a photon collides with an electron, transferring some of its energy to the electron. The kinetic energy of the recoil electron depends on the energy of the incident X-ray photons. The equation for the kinetic energy of the recoil electron in Compton scattering is given by E², where E represents the energy of the incident X-ray photons. As the energy of the photons increases, the kinetic energy of the recoil electron also increases, following a quadratic relationship.

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Matt Balanda |BS, Science |
Physics Expert
Matt graduated with a Master's in Educational Leadership for Faith-Based Schools from California Baptist University and a Bachelor's of Science in Aerospace Engineering and Mathematics from the University of Arizona. A devoted leader, transitioned from Aerospace Engineering to inspire students. As the High School Vice-Principal and a skilled Physics teacher at Calvary Chapel Christian School, his passion is nurturing a love for learning and deepening students' connection with God, fostering a transformative educational journey.

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