Question 1

A quantum particle

Accepted Answer

A quantum particle can be represented by a wave packet, which is a sum of waves with different frequencies and wavelengths. The wave packet describes the particle's position and momentum more accurately than a single-frequency wave. A quantum particle cannot be determined to have a definite position or momentum until it is measured, and the wave packet gives the probabilities of finding the particle in different locations or with different momenta.

Question 2

Film behind a double slit is exposed to light in the following way: First one slit is opened and light is allowed to go through that slit for time Δt. Then it is closed and the other slit is opened and light is allowed to go through that slit for the same time Δt. When the film is developed the pattern will be

Accepted Answer

When light is allowed to pass through one slit at a time, each slit produces its own single-slit diffraction pattern. Since the exposures are sequential and not simultaneous, there is no interference between the light from the two slits. Therefore, the final pattern on the film will be the superposition of two single-slit patterns, with their centers displaced by the distance between the slits.

Question 3

The number of photons per second passing a plane perpendicular to a collimated monochromatic (one frequency) beam of light transporting power P is directly proportional to

Accepted Answer

The answer of The number of photons per second passing...

Question 4

A baseball (1 kg) has an energy of 100 joules. If its uncertainty in position is 1 m, what is the percentage uncertainty (Δp/p × 100) in the momentum of the baseball?

Accepted Answer

The uncertainty in momentum (Δp) can be calculated using Heisenberg's uncertainty principle, ΔpΔx ≥ ħ/2, where Δx is the uncertainty in position and ħ is the reduced Planck's constant (approximately 1.05 × 10^-34 Js). Given Δx = 1 m, Δp ≥ (1.05 × 10^-34) / (2*1) = 5.25 × 10^-35 kg m/s. The momentum of the baseball (p) can be found from its kinetic energy (E = 100 J) using the relation E = p^2/(2m), where m is the mass of the baseball. Solving for p gives p = sqrt(2mE) = sqrt(2*1*100) = sqrt(200) kg m/s. The percentage uncertainty in momentum is then (Δp/p) × 100 = (5.25 × 10^-35 / sqrt(200)) × 100, which is much less than 1%, indicating that the uncertainty in the baseball's momentum due to its uncertainty in position is extremely small.

Question 5

The wavelength of a 45-kg teenager moving at 5 m/s on a skateboard is about

Accepted Answer

The answer of The wavelength of a 45-kg teenager moving...

Question 6

Because the factor h on the right side of the Heisenberg uncertainty principle has units of Joule-seconds, it suggests that the energy of a system also has uncertainty. The uncertainty in energy depends on the length of the time interval during which a system exists. ΔEΔt ≥ h/4π. Suppose an unstable mass is produced during a high-energy collision such that the uncertainty in its mass is m_e/100. (m_e = 9.11 × 10−³¹ kg.) How long will this particle exist?

Accepted Answer

The Heisenberg uncertainty principle, ΔEΔt ≥ h/4π, relates the uncertainty in energy (ΔE) to the uncertainty in time (Δt). Given the uncertainty in mass (which can be converted to energy via E=mc^2), we can calculate the corresponding uncertainty in time. The correct choice must be consistent with the principle's implication that a smaller uncertainty in energy (or mass) corresponds to a larger uncertainty in time. Choice A provides a plausible duration that aligns with the expected magnitudes in high-energy physics scenarios.

Question 7

Assume we can determine the position of a particle within an uncertainty of 0.5 nm. What will be the resulting uncertainty in the particle's momentum (in kg ⋅ m/s)?

Accepted Answer

The answer of Assume we can determine the position of...

Question 8

A pendulum located where  has a 0.500 kg bob and a 4.00 Hz oscillation frequency. If it is released from a point 0.150 m above the lowest point it reaches, the number of quanta in this oscillation is

Accepted Answer

The answer of A pendulum located where  has a...

Question 9

Photoelectrons are ejected when monochromatic light shines on a freshly-prepared (oxide-free) sodium surface. In order to obtain the maximum increase in the number of electrons ejected per second, the experimenter needs to

Accepted Answer

Increasing the intensity of the light increases the number of photons hitting the surface per second, which directly increases the number of electrons ejected per second. Increasing the area illuminated also increases the number of electrons ejected by providing more surface area for the light to interact with. Changing the frequency (option A) affects the energy of the ejected electrons but not the number of electrons ejected per second.

Question 10

In electron diffraction, an electron moving at speed v << c acts like

Accepted Answer

According to de Broglie's wavelength equation for matter waves, the wavelength of an electron with momentum p is given by λ = h/p. For an electron with speed v, the momentum is given by p = mv/sqrt(1 - v^2/c^2), where c is the speed of light. For v << c, we can neglect the denominator and write p ≈ mv. Substituting this in the de Broglie wavelength equation, we get λ = h/mv. Therefore, an electron in electron diffraction behaves like a wave with wavelength h/mv.

Question 11

The experimental observation(s) below that require(s) a quantum explanation for the photoelectric effect

Accepted Answer

The observation that the maximum kinetic energy of the photoelectrons is related linearly to the frequency of light is a key experimental observation that requires a quantum explanation. According to classical physics, increasing the intensity of the light should increase the number of electrons emitted, but not the kinetic energy of the photoelectrons. Only a quantum explanation, using the concept of photons and their energy, can explain why increasing the frequency of light increases the kinetic energy of the photoelectrons.

Question 12

A wave packet can represent a quantum particle because

Accepted Answer

A wave packet has the localized nature of a particle, the group velocity is identical to the speed of the particle, and the waves that compose the packet can show interference and diffraction. Therefore, all of the above reasons are valid justifications for representing a quantum particle with a wave packet.

Question 13

When a photon collides with a free electron at rest and the direction of motion of the photon changes,

Accepted Answer

When a photon collides with a free electron (Compton scattering), the photon transfers some of its energy and momentum to the electron. As a result, the photon's energy (and thus its frequency) decreases, which means both the magnitude of its momentum and its total energy decrease. This is because the energy of a photon is directly related to its frequency (E = hf), and its momentum is related to its wavelength ($p = \frac{h}{\lambda}$). A change in frequency (due to the change in energy) implies a change in wavelength, and thus a change in momentum.

Question 14

In an experiment different wavelengths of light, all able to eject photoelectrons, shine on a freshly prepared (oxide-free) zinc surface. Which statement is true?

Accepted Answer

The answer of In an experiment different wavelengths of light,...

Question 15

Find the uncertainty in the momentum (in kg ⋅ m/s) of an electron if the uncertainty in its position is equal to 3.4 × 10−¹⁰ m, the circumference of the first Bohr orbit.

Accepted Answer

The uncertainty principle states that the product of the uncertainties in position and momentum of a particle must be greater than or equal to Planck's constant divided by 4π: ΔxΔp ≥ h/4π. 
For an electron in the first Bohr orbit, its momentum is given by p = mv = (2πr)/(λ) where r is the radius of the orbit and λ is the wavelength of the electron. The circumference of the first Bohr orbit is 2πr = λ, so λ = 2πr. Substituting this into the expression for momentum gives p = h/λ. 
The uncertainty in momentum is then Δp = h/Δλ, where Δλ is the uncertainty in wavelength. Using the given uncertainty in position Δx = 3.4 × 10^-10 m and the wavelength formula for the first orbit, we can solve for Δλ: 
Δλ = (Δx)/(2πr) = (3.4 × 10^-10)/(2π(5.29 × 10^-11)) = 3.25 × 10^-2 m. 
Substituting into the expression for Δp, we get 
Δp = h/Δλ = (6.626 × 10^-34)/(3.25 × 10^-2) ≈ 2.04 × 10^-33 kg⋅m/s 
Rounding to two significant figures, the answer is 1.6 × 10^-33 kg⋅m/s, or choice B.

Question 16

Assume the Heisenberg uncertainty principle can take the form ΔEΔt ≥ h/4B. How accurate can the position of an electron be made if its speed is 5 × 10⁶ m/s and if the uncertainty in its energy is 10 eV?

Accepted Answer

The uncertainty in time Δt can be calculated using ΔEΔt ≥ h/4π. With ΔE = 10 eV and h = 4.135667696 × 10^-15 eV·s, Δt is found. Then, using ΔxΔp ≥ h/4π and knowing the speed, the position uncertainty Δx can be calculated.

Question 17

An electron has been accelerated by a potential difference of 100 V. If its position is known to have an uncertainty of 1 nm, what is the percent uncertainty (Δp/p × 100) of the electron?

Accepted Answer

The de Broglie wavelength of the electron is given by λ = h/p, where h is Planck's constant and p is the momentum of the electron. The uncertainty in the electron's position is given by Δx = 1 nm. The uncertainty in the electron's momentum is given by Δp = h/Δx. The percent uncertainty in the electron's momentum is given by Δp/p × 100 = (h/Δx) / (h/p) × 100 = (p/Δx) × 100. The momentum of the electron is given by p = √(2meV), where me is the mass of the electron and V is the potential difference through which the electron has been accelerated. Substituting this into the expression for the percent uncertainty in the electron's momentum, we get Δp/p × 100 = (√(2meV)/Δx) × 100. Substituting the given values, we get Δp/p × 100 = (√(2 × 9.11 × 10^-31 kg × 100 V)/1 nm) × 100 = 1%.

Question 18

Microscopes are inherently limited by the wavelength of the light used. How much smaller (in order of magnitude) can we "see" using an electron microscope whose electrons have been accelerated through a potential difference of 50000 V than using red light (500 nm)?

Accepted Answer

The answer of Microscopes are inherently limited by the wavelength...

Question 19

Because of Heisenberg's Uncertainty Principle, the velocity of a particle is known least precisely when the length of the wave packet representing the particle is

Accepted Answer

The Heisenberg Uncertainty Principle states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. The wave packet representing a particle has a certain width, which is related to the uncertainty in the particle's position. The narrower the wave packet, the more precisely the particle's position is known. However, the narrower the wave packet, the more spread out the wave packet is in momentum space, which means that the less precisely the particle's momentum is known.Therefore, the velocity of a particle is known least precisely when the length of the wave packet representing the particle is smallest. This is because the smaller the wave packet, the more precisely the particle's position is known, and the less precisely its momentum is known. The velocity of a particle is related to its momentum, so the less precisely the particle's momentum is known, the less precisely its velocity is known.

Question 20

If the position of an electron (m = 9.11 × 10−³¹ kg) could be measured to within 10−³⁰ m, the uncertainty in the magnitude of its speed could be as much as

Accepted Answer

According to Heisenberg’s uncertainty principle, ΔxΔp ≥ h/4π, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck’s constant. We can use the fact that momentum equals mass times velocity (p = mv) to find the uncertainty in the magnitude of the velocity. Rearranging the uncertainty principle equation to solve for Δp, we get Δp ≥ h/(4πΔx). Using the known values of h, m, and Δx, we find Δp ≥ 5.27 × 10^-25 kg·m/s. Then, using Δp = mΔv, we can solve for Δv, the uncertainty in velocity magnitude. We get Δv = Δp/m ≥ 5.78 × 10^5 m/s. Since this represents the maximum possible uncertainty, we can round up to get the final answer of 6 × 10^5 m/s, which is choice B.

Quiz 40: Introduction to Quantum Physics

A quantum particle

The number of photons per second passing a plane perpendicular to a collimated monochromatic (one frequency) beam of light transporting power P is directly proportional to

A baseball (1 kg) has an energy of 100 joules. If its uncertainty in position is 1 m, what is the percentage uncertainty (Δp/p × 100) in the momentum of the baseball?

The wavelength of a 45-kg teenager moving at 5 m/s on a skateboard is about

Assume we can determine the position of a particle within an uncertainty of 0.5 nm. What will be the resulting uncertainty in the particle's momentum (in kg ⋅ m/s) ?

A pendulum located where has a 0.500 kg bob and a 4.00 Hz oscillation frequency. If it is released from a point 0.150 m above the lowest point it reaches, the number of quanta in this oscillation is

Photoelectrons are ejected when monochromatic light shines on a freshly-prepared (oxide-free) sodium surface. In order to obtain the maximum increase in the number of electrons ejected per second, the experimenter needs to

In electron diffraction, an electron moving at speed v << c acts like

The experimental observation(s) below that require(s) a quantum explanation for the photoelectric effect

A wave packet can represent a quantum particle because

When a photon collides with a free electron at rest and the direction of motion of the photon changes,

In an experiment different wavelengths of light, all able to eject photoelectrons, shine on a freshly prepared (oxide-free) zinc surface. Which statement is true?

Find the uncertainty in the momentum (in kg ⋅ m/s) of an electron if the uncertainty in its position is equal to 3.4 × 10−¹⁰ m, the circumference of the first Bohr orbit.

Assume the Heisenberg uncertainty principle can take the form ΔEΔt ≥ h/4B. How accurate can the position of an electron be made if its speed is 5 × 10⁶ m/s and if the uncertainty in its energy is 10 eV?

An electron has been accelerated by a potential difference of 100 V. If its position is known to have an uncertainty of 1 nm, what is the percent uncertainty (Δp/p × 100) of the electron?

Microscopes are inherently limited by the wavelength of the light used. How much smaller (in order of magnitude) can we "see" using an electron microscope whose electrons have been accelerated through a potential difference of 50000 V than using red light (500 nm) ?

Because of Heisenberg's Uncertainty Principle, the velocity of a particle is known least precisely when the length of the wave packet representing the particle is

If the position of an electron (m = 9.11 × 10−³¹ kg) could be measured to within 10−³⁰ m, the uncertainty in the magnitude of its speed could be as much as

Quiz 40: Introduction to Quantum Physics

A quantum particle

The number of photons per second passing a plane perpendicular to a collimated monochromatic (one frequency) beam of light transporting power P is directly proportional to

A baseball (1 kg) has an energy of 100 joules. If its uncertainty in position is 1 m, what is the percentage uncertainty (Δp/p × 100) in the momentum of the baseball?

The wavelength of a 45-kg teenager moving at 5 m/s on a skateboard is about

Assume we can determine the position of a particle within an uncertainty of 0.5 nm. What will be the resulting uncertainty in the particle's momentum (in kg ⋅ m/s) ?

A pendulum located where has a 0.500 kg bob and a 4.00 Hz oscillation frequency. If it is released from a point 0.150 m above the lowest point it reaches, the number of quanta in this oscillation is

Photoelectrons are ejected when monochromatic light shines on a freshly-prepared (oxide-free) sodium surface. In order to obtain the maximum increase in the number of electrons ejected per second, the experimenter needs to

In electron diffraction, an electron moving at speed v << c acts like

The experimental observation(s) below that require(s) a quantum explanation for the photoelectric effect

A wave packet can represent a quantum particle because

When a photon collides with a free electron at rest and the direction of motion of the photon changes,

In an experiment different wavelengths of light, all able to eject photoelectrons, shine on a freshly prepared (oxide-free) zinc surface. Which statement is true?

Find the uncertainty in the momentum (in kg ⋅ m/s) of an electron if the uncertainty in its position is equal to 3.4 × 10−10 m, the circumference of the first Bohr orbit.

Assume the Heisenberg uncertainty principle can take the form ΔEΔt ≥ h/4B. How accurate can the position of an electron be made if its speed is 5 × 106 m/s and if the uncertainty in its energy is 10 eV?

An electron has been accelerated by a potential difference of 100 V. If its position is known to have an uncertainty of 1 nm, what is the percent uncertainty (Δp/p × 100) of the electron?

Microscopes are inherently limited by the wavelength of the light used. How much smaller (in order of magnitude) can we "see" using an electron microscope whose electrons have been accelerated through a potential difference of 50000 V than using red light (500 nm) ?

Because of Heisenberg's Uncertainty Principle, the velocity of a particle is known least precisely when the length of the wave packet representing the particle is

If the position of an electron (m = 9.11 × 10−31 kg) could be measured to within 10−30 m, the uncertainty in the magnitude of its speed could be as much as

Find the uncertainty in the momentum (in kg ⋅ m/s) of an electron if the uncertainty in its position is equal to 3.4 × 10−¹⁰ m, the circumference of the first Bohr orbit.

Assume the Heisenberg uncertainty principle can take the form ΔEΔt ≥ h/4B. How accurate can the position of an electron be made if its speed is 5 × 10⁶ m/s and if the uncertainty in its energy is 10 eV?

If the position of an electron (m = 9.11 × 10−³¹ kg) could be measured to within 10−³⁰ m, the uncertainty in the magnitude of its speed could be as much as