Question 1

What is the longest wavelength of light that can cause photoelectron emission from a metal that has a work function of 2.20 eV?

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Question 2

What is the momentum of a photon of light that has a wavelength of 480 nm?

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Question 3

A certain photon after being scattered from a free electron at rest moves at an angle of 120° with respect to the incident direction. If the wavelength of the incident photon is 0.591 nm, what is the energy of the scattered photon?

Accepted Answer

We can use the Compton scattering formula: $\Delta\lambda=\dfrac{h}{mc}(1-\cos heta)$ where $\Delta\lambda$ is the change in wavelength, $h$ is Planck's constant, $m$ is the mass of the electron, $c$ is the speed of light, and $ heta$ is the scattering angle. We can re-arrange the formula to solve for the change in wavelength: $\Delta\lambda=\lambda'-\lambda=\dfrac{h}{mc}(1-\cos heta)$ where $\lambda'$ is the wavelength of the scattered photon and $\lambda$ is the wavelength of the incident photon. Plugging in the given values, we get: $\Delta\lambda=\lambda'-\lambda=\dfrac{h}{mc}(1-\cos heta)=\dfrac{6.626 imes10^{-34} ext{ J}\cdot ext{s}}{9.109 imes10^{-31} ext{ kg}\cdot 3 imes10^8 ext{ m/s}}(1-\cos 120^\circ)=2.42 imes10^{-12} ext{ m}$ We can then use the formula for energy of a photon: $E=hf=\dfrac{hc}{\lambda}$ where $f$ is the frequency of the photon. We can solve for the energy of the scattered photon by using the change in wavelength: $E'=\dfrac{hc}{\lambda'}=\dfrac{hc}{\lambda-\Delta\lambda}=\dfrac{hc}{0.591 imes10^{-9} ext{ m}-2.42 imes10^{-12} ext{ m}}=3.34 imes10^{-16} ext{ J}$ Therefore, the answer is D) 3.34 × 10^-16 J.

Question 4

A beam of white light of a certain frequency is incident on a metal surface. If the work function of this surface is 2.20 eV, what frequency of light would give a maximum kinetic energy of 0.25 eV to the photoelectrons ejected from this surface?

Accepted Answer

The energy of the incident light must be equal to the work function plus the maximum kinetic energy of the photoelectrons. Using the equation E = hf, where h is Planck's constant (4.135667696 × 10^-15 eV·s), we find the frequency: f = (2.20 eV + 0.25 eV) / (4.135667696 × 10^-15 eV·s) = 5.92 × 10^14 Hz.

Question 5

What is the minimum uncertainty in determining the position of an electron if the speed of the electron is 6000 m/s within an accuracy of 0.004%?

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Question 6

A beam of light with a frequency range of 3.01 × 10¹⁴ Hz - 6.10 × 10¹⁴ Hz is incident on a metal surface. If the work function of the metal surface is 2.20 eV, what is the range of frequencies in this beam of light that will be responsible for ejecting electrons from this metal surface?

Accepted Answer

The threshold frequency (f₀) can be calculated using the work function (φ) and Planck's constant (h): f₀ = φ/h. For φ = 2.20 eV, f₀ ≈ 5.31 × 10¹⁴ Hz. Frequencies above this will eject electrons.

Question 7

A certain crystal has an interatomic spacing of 0.442 nm. A beam of neutrons moving with a speed of 1640 m/s is diffracted from the crystal. What is the de Broglie wavelength of the neutron?

Accepted Answer

The de Broglie wavelength of a particle is given by: 
λ = h/p
where h is Planck's constant and p is the momentum of the particle. For a neutron, the momentum is given by:
p = mv
where m is the mass of the neutron and v is its velocity.

First, we need to convert the given velocity of the neutron into momentum: 
p = mv = (1.6749 × 10^-27 kg)(1640 m/s) = 2.74996 × 10^-22 kg m/s

Then, we can use the formula for de Broglie wavelength: 
λ = h/p = (6.626 × 10^-34 J s)/(2.74996 × 10^-22 kg m/s) ≈ 0.241 nm

Therefore the answer is C, 0.241 nm.

Question 8

A certain photon after being scattered from a free electron at rest moves at an angle of 120° with respect to the incident direction. If the wavelength of the incident photon is 0.611 nm, what is the kinetic energy of the recoiling electron?

Accepted Answer

The scattering of a photon by a free electron can be modeled as an elastic collision between the photon and the electron. In an elastic collision, energy is conserved, so the total energy of the system before the collision is equal to the total energy after the collision. The energy of a photon can be related to its wavelength by the equation:

E = hc/λ

where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

In this problem, we are given the wavelength of the incident photon (λ = 0.611 nm) and the angle by which it is scattered (θ = 120°). From the law of conservation of momentum, we know that the momentum of the recoiling electron must be equal and opposite to the momentum of the scattered photon. We can use this fact, along with the conservation of energy, to solve for the kinetic energy of the electron.

The momentum of a photon is given by:

p = h/λ

where p is the momentum of the photon. The momentum of the scattered photon can be related to the momentum of the recoiling electron by the equation:

p = mev

where me is the mass of the electron, v is its velocity, and p and v are in the same direction.

By conservation of momentum, we have:

p = 2p sin(θ/2)

where θ is the scattering angle. Substituting in the expressions for p and mev above, we can solve for v:

v = p/me = (2h/λme) sin(θ/2)

Once we have the velocity of the recoiling electron, we can use the equation for kinetic energy:

KE = (1/2)mev^2

to solve for its kinetic energy.

Plugging in the given values, we get:

KE = (1/2)(9.109 × 10^-31 kg)(3.00 × 10^8 m/s)(2π/h)(0.611 × 10^-9 m)^2 sin^2(60°)

KE = 2.1 × 10^-18 J

Therefore, the answer is B.

Question 9

What is the momentum of a photon of light that has a frequency of 2.8 × 10¹⁴ Hz?

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Question 10

Light with a frequency of 8.70 × 10¹⁴ Hz is incident on a metal that has a work function of 2.8 eV. What is the maximum kinetic energy that a photoelectron ejected in this process can have?

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Question 11

A certain photon after being scattered from a free electron at rest moves at an angle of 120° with respect to the incident direction. If the wavelength of the incident photon is 0.591 nm, what is the wavelength of the scattered photon?

Accepted Answer

The change in wavelength due to Compton scattering is given by the formula: Δλ = (h/mc) * (1 - cos(θ)), where h is Planck's constant, m is the electron rest mass, c is the speed of light, and θ is the scattering angle. For θ = 120°, Δλ is approximately 0.004 nm. Adding this to the initial wavelength of 0.591 nm gives approximately 0.595 nm.

Question 12

A beam of light with a frequency range of 3.01 × 10¹⁴ Hz - 6.10 × 10¹⁴ Hz is incident on a metal surface. If the work function of the metal surface is 2.20 eV, what is the maximum kinetic energy of photoelectrons ejected from this surface?

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Question 13

If the momentum of an electron is 1.95 × 10^-27 kg∙m/s, what is its de Broglie wavelength?

Accepted Answer

The de Broglie wavelength (λ) is given by the formula:
λ = h/p
Where h is the Planck's constant and p is the momentum of the particle.

Substituting the given values, we get:
λ = (6.626 × 10^-34 J∙s)/(1.95 × 10^-27 kg∙m/s) 
λ = 3.40 × 10^-7 m 
λ = 340 nm

Therefore, the correct answer is A) 340 nm.

Question 14

An electron is moving with the speed of 1780 m/s. What is its de Broglie wavelength?

Accepted Answer

The de Broglie wavelength is calculated using the formula λ = h / (mv), where h is Planck's constant (6.626 x 10^-34 m^2 kg/s), m is the mass of an electron (9.109 x 10^-31 kg), and v is the velocity (1780 m/s). Plugging in these values gives λ ≈ 409 nm.

Question 15

What is the uncertainty in determining the momentum of an electron if the speed of the electron is 6000 m/s within an accuracy of 0.004%?

Accepted Answer

The uncertainty in momentum can be found using the formula:

Δp = mΔv

where Δv is the uncertainty in velocity and m is the mass of the electron.

Δv can be found by multiplying the speed of the electron by its accuracy:

Δv = (6000 m/s) x (0.00004) = 0.24 m/s

The mass of an electron is approximately 9.11 x 10^-31 kg.

Δp = (9.11 x 10^-31 kg) x (0.24 m/s) = 2.19 x 10^-31 kg∙m/s

Therefore, the uncertainty in determining the momentum of an electron with a speed of 6000 m/s within an accuracy of 0.004% is 2.19 x 10^-31 kg∙m/s.

The correct answer is C.

Question 16

A certain crystal has an interatomic spacing of 0.142 nm. If a beam of electrons moving with a speed of 1400 m/s is diffracted from this crystal, what is the de Broglie wavelength of the electron?

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Question 17

The de Broglie wavelength of an electron is 380 nm. What is the speed of this electron?

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Question 18

Light with a wavelength of 310 nm is incident on a metal that has a work function of 3.8 eV. What is the maximum kinetic energy that a photoelectron ejected in this process can have?

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Question 19

A beam of X-rays at a certain wavelength are scattered from a free electron at rest and the scattered beam is observed at 45.0° to the incident beam. What is the Compton shift?

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Question 20

A certain crystal has an interatomic spacing of 0.442 nm. A beam of neutrons moving with a speed of 1640 m/s is diffracted from the crystal. What is the angle of the second interference maximum?

Accepted Answer

The de Broglie wavelength of the neutrons is calculated using λ = h/(mv), where h is Planck's constant and mv is the momentum. Using the Bragg's law nλ = 2d sin(θ) for the second maximum (n=2), we solve for θ.

Quiz 30: Quantum Physics