Physics Test: Practice Questions On Compton Effect!

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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.

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

The correct answer is "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 silver.

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.

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.

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.

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.

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.

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.