Quantum Physics

விளையாடிக் கற்றுக்கொள்ளுங்கள்

ஆற்றல் சம்பாதிக்க இந்த கேள்விகளுக்குப் பதிலளியுங்கள், பின்னர் மீன் பிடித்து ஆராயுங்கள். கணக்கு தேவையில்லை.

கல்வியாளர்களுக்கு: Quantum Physics (Physics, HL)-க்கான தயாரான பாட ஸ்லைடுகள், திருப்புதல் குறிப்புகள் — உங்கள் பாடத்தில் அவற்றைப் பயன்படுத்தவும், அல்லது கற்பவர்கள் நேரலை விளையாட்டாக விளையாடும் ஊடாடும் வகுப்பு செயல்பாடாக தலைப்பை இயக்கவும்.

பாட குறிப்புகள்

The Photoelectric Effect

The photoelectric effect is the emission of electrons from a metal surface when it absorbs electromagnetic radiation.
The emitted electrons are called photoelectrons.
It provides evidence that light behaves as a particle (is quantised) because each electron absorbs only a single photon.
The threshold frequency is the minimum frequency of incident radiation required to remove a photoelectron from the surface of a metal.
The work function Φ is the minimum energy required to release a photoelectron from the surface of a metal.
Threshold frequency and work function are properties of the metal and vary from metal to metal.
Alkali metals (e.g., sodium, potassium) have threshold frequencies in the visible region; transition metals (e.g., zinc, iron) have threshold frequencies in the ultraviolet region.

The Photoelectric Equation

The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency.
The photoelectric equation is hf = Φ + E_k max.
E_k max is the maximum kinetic energy of an emitted photoelectron.
The equation can be rearranged as E_k max = hf − Φ.
Photons with energy less than the work function do not liberate photoelectrons.
Photons with frequency equal to the threshold frequency f₀ have energy hf₀ = Φ.
A graph of E_k max against f is a straight line with gradient h (Planck's constant), y-intercept −Φ, and x-intercept f₀.

Intensity, Current and Stopping Potential

The kinetic energy of photoelectrons depends only on the frequency of the incident radiation, not on its intensity.
Increasing the intensity increases the number of photons incident per second, hence the number of photoelectrons emitted per second (the photoelectric current).
The photoelectric current is proportional to the intensity of the incident radiation.
The stopping potential V_s is the potential difference required to stop photoelectron emission.
At the stopping potential, the energy needed to cross the gap equals the maximum kinetic energy: E_k max = eV_s.
Increasing intensity increases the photoelectric current but the stopping potential remains the same, confirming intensity does not affect kinetic energy.
If frequency is increased while intensity is constant, the photoelectric current decreases because fewer photons are incident per second.

The Particle Nature of Light

The photoelectric effect is evidence for the particle nature of light.
The wave theory predicts that photoelectrons should be released at any frequency and that kinetic energy should depend on intensity, which is not observed.
Observations from the gold-leaf electroscope experiment: the leaf falls instantly when illuminated with UV light.
At lower intensities, the leaf falls more slowly but still instantly, supporting one-to-one photon-electron interactions.
The leaf does not fall at lower frequencies, even if intensity is increased, supporting the existence of a threshold frequency and work function.
Einstein proposed that light consists of quanta of energy called photons, with energy E = hf.

The de Broglie Wavelength

Louis de Broglie hypothesised that all moving particles have an associated matter wave.
The de Broglie wavelength λ is given by λ = h/p, where p is momentum.
Since p = mv, the equation can be written as λ = h/(mv).
The de Broglie wavelength can also be related to kinetic energy: λ = h/√(2mE_k).
For everyday objects, the de Broglie wavelength is too small to observe quantum effects.
Electron diffraction through a graphite film provides evidence for the wave nature of matter.
When electrons are accelerated through a larger potential difference, their speed and momentum increase, so their de Broglie wavelength decreases, leading to a smaller diffraction pattern.

Wave-Particle Duality

Light exhibits wave-particle duality: it behaves as both a particle and a wave.
Light interacts with matter as a particle (photoelectric effect).
Light propagates through space as a wave (diffraction and interference in Young's double-slit experiment).
The wave theory fails to explain the photoelectric effect, particularly the threshold frequency and the independence of kinetic energy from intensity.
The photon model explains that electromagnetic waves carry energy in discrete packets called photons.
The development of wave-particle duality resolved a 300-year debate about the nature of light.

Compton Scattering

Compton scattering is the interaction of a high-energy photon (X-ray or gamma) with an orbital electron, causing an increase in the photon's wavelength and ejection of the electron.
It provides further evidence for the particle nature of light.
During the collision, the photon transfers energy to the electron, so the photon's energy decreases and its wavelength increases.
The electron and photon are deflected in different directions due to conservation of momentum.
The Compton formula is Δλ = (h/m_e c)(1 − cos θ), where θ is the scattering angle.
The constant h/m_e c is the Compton wavelength.
The change in wavelength is zero at θ = 0°, and maximum at θ = 180° (twice the Compton wavelength).

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பதிவு செய்க — இலவசம்உள்நுழைக

பயிற்சி கேள்விகள்

இலவச முன்னோட்டம் — 62-இல் 8 கேள்விகள். அனைத்தையும் பார்க்க பதிவு செய்யவும்.

1.Which of the following is evidence for the wave nature of electrons?
Easy
- AElectron diffraction from crystals
- BThe photoelectric effect
- CContinuous energy spectrum in β decay
- DPair production
2.Which of the following experiments provides evidence for the particle nature of light?
Easy
- AGamma decay
- BYoung's double slit experiment
- CThe photoelectric effect
- DElectron diffraction
3.According to the de Broglie equation, the wavelength of an electron is:
Easy
- Adirectly proportional to the mass of the electron
- Binversely proportional to the square of the velocity of the electron
- Cdirectly proportional to the velocity of the electron
- Dinversely proportional to the mass of the electron
4.The work functions of four metals are given: sodium 2.3 eV, caesium 2.1 eV, calcium 2.9 eV, silver 4.3 eV. Which metal will exhibit the photoelectric effect most readily?
Easy
- ASodium
- BCaesium
- CCalcium
- DSilver
5.The photoelectric effect provides evidence that light behaves as a particle.
Easy
True or false?
6.The maximum kinetic energy of emitted photoelectrons depends on the intensity of the incident radiation.
Easy
True or false?
7.Which of the following statements about the photoelectric effect are correct? (select all that apply)
Medium
- APhotoelectrons are emitted only if the incident light frequency is above the threshold frequency
- BIncreasing the intensity of incident light increases the maximum kinetic energy of photoelectrons
- CThe photoelectric current is proportional to the intensity of the incident radiation
- DThe work function is the minimum energy required to release a photoelectron from the metal surface
- EPhotoelectrons are emitted with a range of kinetic energies up to a maximum value
8.Match each term with its correct definition.
Easy
- Photon
- Work function
- Threshold frequency
- Stopping potential
- A quantum of electromagnetic energy
- The minimum energy required to release a photoelectron from a metal surface
- The minimum frequency of incident radiation required to remove a photoelectron
- The potential difference required to stop photoelectron emission

மேலும் கேள்விகள்

9.Which graph shows how intensity I varies with angle θ when electrons are diffracted by a nucleus?
Easy
- AA graph with a central peak and decreasing side peaks
- BA graph with uniform intensity at all angles
- CA graph with a single sharp peak at θ = 0°
- DA graph with two symmetric peaks at ±90°
10.Place the steps of the photoelectric effect in the correct order from photon absorption to electron emission.
Medium
- A photon is absorbed by an electron in the metal.
- The electron gains energy equal to the photon energy hf.
- The electron uses energy equal to the work function to escape the metal surface.
- The electron is emitted with maximum kinetic energy Ek max = hf - Φ.
11.The work function of a metal is the minimum energy required to release a photoelectron from its surface.
Easy
True or false?
12.A metal surface is illuminated with light of frequency below its threshold frequency. What happens to the emission of photoelectrons if the intensity of the light is increased?
Medium
- ANo photoelectrons are emitted.
- BMore photoelectrons are emitted.
- CPhotoelectrons are emitted with greater kinetic energy.
- DPhotoelectrons are emitted after a delay.
13.In a Compton scattering event, an X-ray photon scatters through an angle θ. Which of the following correctly describes the change in wavelength Δλ of the photon?
Medium
- AΔλ = (h/m_e c)(1 - cos θ)
- BΔλ = (h/m_e c)(1 + cos θ)
- CΔλ = (h/m_e c)(cos θ - 1)
- DΔλ = (h/m_e c) sin θ
14.Which of the following are evidence for the wave nature of electrons? (select all that apply)
Medium
- AElectron diffraction from crystals
- BThe photoelectric effect
- CYoung's double slit experiment with electrons
- DCompton scattering
- EElectron interference patterns
15.Match each quantity with its correct relationship in the photoelectric effect.
Medium
- Maximum kinetic energy
- Photoelectric current
- Stopping potential
- Threshold frequency
- Depends only on frequency of incident light and work function
- Proportional to intensity of incident light
- Related to maximum kinetic energy by Ek max = eVs
- Minimum frequency required for photoelectric emission
16.Order the following steps to show how the de Broglie wavelength of electrons changes as accelerating voltage increases in an electron diffraction tube.
Hard
- Accelerating voltage is increased.
- Kinetic energy of electrons increases.
- Momentum of electrons increases.
- De Broglie wavelength decreases.
- Radius of diffraction rings decreases.
17.In a photoelectric experiment, monochromatic light of frequency f illuminates a metal cathode with threshold frequency f0. The stopping potential V is measured. Which equation correctly relates V, f, and f0?
Medium
- AeV = h(f - f0)
- BeV = h(f + f0)
- CeV = hf0 - hf
- DeV = hf/f0
18.Which of the following statements about the gold leaf electroscope experiment are correct? (select all that apply)
Medium
- AThe gold leaf falls instantly when illuminated with UV light.
- BThe gold leaf falls more slowly at lower intensities of UV light.
- CThe gold leaf falls when illuminated with visible light of high intensity.
- DThe gold leaf does not fall below a certain frequency even if intensity is increased.
- EThe gold leaf falls because it becomes more negatively charged.
19.In the photoelectric effect, increasing the intensity of incident light increases the number of photoelectrons emitted per second.
Easy
True or false?
20.Place the following types of electromagnetic radiation in order of increasing photon energy.
Medium
- Infrared
- Visible light
- Ultraviolet
- X-ray
- Gamma ray
21.A photon of energy E and wavelength λ strikes a stationary electron and transfers a fraction of its energy as it scatters. The scattered photon has a wavelength λ'. What is the speed of the recoiling electron?
Medium
- A√( (2E/m_e) (λ' - λ)/λ' )
- B√( (2E/m_e) (λ' - λ)/λ )
- C√( (2E/m_e) (λ - λ')/λ )
- D√( (2E/m_e) (λ'/λ) )
22.Which of the following correctly describe the photoelectric effect? (select all that apply)
Medium
- AElectrons are emitted from a metal surface when light is absorbed.
- BThe emitted electrons are called photoelectrons.
- CThe effect provides evidence for the wave nature of light.
- DThe maximum kinetic energy of photoelectrons increases with frequency of incident light.
- EThe work function is the minimum energy to remove an electron from the metal.
23.The stopping potential in a photoelectric experiment is independent of the intensity of the incident light.
Medium
True or false?
24.Match each graph feature with the quantity it represents for a graph of maximum kinetic energy E_k against frequency f.
Medium
- Gradient
- y-intercept
- x-intercept
- Value at f = 0
- Planck's constant h
- Work function Φ (negative)
- Threshold frequency f0
- No electrons emitted
25.Order the steps to calculate the threshold frequency of a metal given its work function.
Hard
- Identify the work function Φ in eV.
- Convert Φ to joules using 1 eV = 1.6 × 10^-19 J.
- Use the equation Φ = hf0.
- Rearrange to f0 = Φ/h.
- Substitute values and calculate f0.
26.An X-ray photon scatters from a stationary electron. The scattered photon has a longer wavelength than the incident photon. Which of the following statements best explains this?
Medium
- AThe photon transfers some of its energy to the electron, so its energy and therefore frequency decrease, increasing its wavelength.
- BThe photon loses speed as it collides with the electron, so its wavelength increases.
- CThe electron absorbs the photon and then re-emits a photon of lower energy after a time delay.
- DThe photon's momentum is conserved, so its wavelength must increase to compensate.
27.In a photoelectric experiment, the stopping potential is measured for light of two different wavelengths but the same intensity. The graph of stopping potential V against 1/λ is a straight line. What does the gradient of this line represent?
Medium
- Ahc/e
- Bh/e
- CΦ/e
- De/hc
28.A beam of electrons is accelerated from rest through a potential difference V and diffracted by a crystal with atomic spacing d. The first-order diffraction angle θ satisfies sin θ = λ/d. If the accelerating voltage is increased by a factor of 4, what happens to sin θ?
Hard
- AIt is halved.
- BIt is doubled.
- CIt is quartered.
- DIt remains unchanged.
29.In the photoelectric effect, increasing the intensity of incident light increases the maximum kinetic energy of the emitted photoelectrons.
Easy
True or false?
30.Place the following steps in the correct order to show how the radius of the diffraction rings changes as the accelerating voltage in an electron diffraction tube is increased.
Medium
- The accelerating voltage is increased.
- The speed and momentum of the electrons increase.
- The de Broglie wavelength of the electrons decreases.
- The angle of diffraction decreases.
- The radius of the diffraction rings decreases.
31.In a Compton scattering experiment, X-rays of wavelength 0.0200 nm are scattered through an angle of 90°. The scattered X-rays have a wavelength of 0.0224 nm. What is the kinetic energy of the recoiling electron? (Use hc = 1.24 × 10³ eV nm.)
Challenge
- A6.9 keV
- B55 keV
- C0.69 keV
- D12 keV
32.Which of the following equations correctly relates the stopping potential V_s to the maximum kinetic energy E_k max of photoelectrons?
Medium
- AE_k max = eV_s
- BE_k max = e / V_s
- CE_k max = V_s / e
- DE_k max = hV_s
33.In a photoelectric experiment, the intensity of monochromatic light above the threshold frequency is increased while the frequency is kept constant. Which of the following statements are correct? (select all that apply)
Medium
- AThe photoelectric current increases.
- BThe maximum kinetic energy of the photoelectrons increases.
- CThe number of photoelectrons emitted per second increases.
- DThe stopping potential increases.
- EThe energy of each incident photon increases.
34.A graph of maximum kinetic energy E_k max against frequency f for a metal is a straight line. What does the gradient of this graph represent?
Medium
- APlanck's constant h
- BThe work function ϕ
- CThe threshold frequency f₀
- DThe electron charge e
35.A student sets up a gold-leaf electroscope with a zinc plate on top. The plate is given a negative charge, so the gold leaf deflects. When ultraviolet light is shone on the plate, the leaf falls. Which observation would show that the intensity of the UV light affects the rate of photoelectric emission?
Medium
- AThe leaf falls more slowly at lower intensities of UV light
- BThe leaf falls instantly at all intensities of UV light
- CThe leaf does not fall at lower frequencies of light
- DThe leaf falls at the same rate regardless of intensity
36.In a photoelectric experiment, the stopping potential is measured for light of two different wavelengths but the same intensity. The graph of stopping potential V_s (y-axis) against 1/λ (x-axis) is a straight line. What does the gradient of this graph represent?
Medium
- Ahc/e
- Bh/e
- Ce/hc
- Dhc
37.In the gold-leaf electroscope experiment, the gold leaf falls instantly when illuminated by UV light, even at low intensities.
Easy
True or false?
38.In a vacuum photocell, monochromatic light of frequency f illuminates a cathode with threshold frequency f₀. The potential difference across the photocell is increased until the ammeter reads zero. Which of the following equations correctly relate the stopping potential V to f and f₀? (select all that apply)
Hard
- AV = h/e (f − f₀)
- Bf₀ = (hf − eV)/h
- Ch = eV/(f − f₀)
- DV = e/h (f − f₀)
- EV = (hf₀ − hf)/e
39.Match each experimental observation from the gold-leaf electroscope experiment with its correct explanation.
Medium
- Gold leaf falls instantly when UV light is shone on the plate
- Gold leaf falls more slowly at lower intensities of UV light
- Gold leaf does not fall when lower-frequency light is used, even at high intensity
- One photon transfers its energy to one electron, so emission is immediate
- Fewer photons strike the plate per second, so fewer photoelectrons are emitted per second
- Each photon has insufficient energy to overcome the work function
40.Place the following steps in the correct order to describe how the radius of the diffraction rings changes as the accelerating voltage in an electron diffraction tube is increased.
Medium
- The accelerating voltage is increased
- The speed of the electrons increases
- The momentum of the electrons increases
- The de Broglie wavelength of the electrons decreases
- The angle of diffraction decreases, so the radius of the rings decreases
41.Complete the photoelectric equation by filling in the blanks.
Easy
hf = ____ + ____
42.What name is given to the electrons emitted from a metal surface when electromagnetic radiation is absorbed?
Easy
43.The work function of sodium is 2.28 eV. Calculate the threshold frequency of sodium in Hz, using h = 6.63 × 10⁻³⁴ J s and 1 eV = 1.6 × 10⁻¹⁹ J. Give your answer to 2 significant figures.
Medium
44.Calculate the de Broglie wavelength of an electron travelling at a speed of 3.18 × 10⁷ m s^-1. Use h = 6.63 × 10^-34 J s and electron mass = 9.11 × 10^-31 kg. Give your answer in metres to 2 significant figures.
Medium
45.An X-ray photon with initial wavelength 7.09 × 10^-11 m scatters through an angle of 45°. Calculate the change in photon wavelength. Use h = 6.63 × 10^-34 J s, m_e = 9.11 × 10^-31 kg, c = 3.00 × 10⁸ m s^-1. Give your answer in metres to 2 significant figures.
Medium
46.Complete the statement about the de Broglie wavelength.
Medium
The de Broglie wavelength is given by λ = ____ / ____, where h is Planck's constant and p is the ____ of the particle.
47.What is the name of the phenomenon observed when electrons are fired at a graphite film and produce a pattern of concentric rings?
Easy
48.Electrons are accelerated through a potential difference of 40 V and diffracted by graphite with atomic spacing 2.1 × 10^-10 m. Calculate the angle of the first minimum of the diffraction pattern, in degrees, to 2 significant figures. Use h = 6.63 × 10^-34 J s, m_e = 9.11 × 10^-31 kg, e = 1.6 × 10^-19 C.
Hard
49.Complete the Compton scattering formula.
Medium
Δλ = (h / (m____e c)) (1 - ____)
50.What is the name of the constant h/m_e c in the Compton scattering formula?
Easy
51.The work function of sodium is 2.28 eV. Calculate the threshold frequency of sodium in Hz. Use h = 6.63 × 10^-34 J s and 1 eV = 1.6 × 10^-19 J. Give your answer to 3 significant figures.
Medium
52.Complete the statement about the photoelectric equation.
Medium
The photoelectric equation is hf = ____ + ____, where the first term is the work function and the second is the maximum kinetic energy.
53.State what is meant by the term photon.
Medium
54.Outline why the wave theory of light cannot explain the observation that no photoelectrons are emitted when the incident light is below a certain frequency, even at high intensity.
Medium
55.Explain how the photoelectric effect provides evidence for the particle nature of light.
Hard
56.Outline why the wave theory of light cannot explain the observation that no photoelectrons are emitted when the incident light is below a certain frequency, even if the intensity is increased.
Hard
57.Explain why the scattered photon in Compton scattering has a longer wavelength than the incident photon.
Hard
58.The work function of sodium is 2.28 eV. Calculate the threshold frequency of sodium in Hz, using h = 6.63 × 10⁻³⁴ J s and 1 eV = 1.6 × 10⁻¹⁹ J. Give your answer to 2 significant figures.
Medium
- A495000000000000 Hz
- B660000000000000 Hz
- C550000000000000 Hz
- D605000000000000 Hz
59.Calculate the de Broglie wavelength of an electron travelling at a speed of 3.18 × 10⁷ m s^-1. Use h = 6.63 × 10^-34 J s and electron mass = 9.11 × 10^-31 kg. Give your answer in metres to 2 significant figures.
Medium
- A2 m
- B1 m
- C-1 m
- D2.3e-11 m
60.An X-ray photon with initial wavelength 7.09 × 10^-11 m scatters through an angle of 45°. Calculate the change in photon wavelength. Use h = 6.63 × 10^-34 J s, m_e = 9.11 × 10^-31 kg, c = 3.00 × 10⁸ m s^-1. Give your answer in metres to 2 significant figures.
Medium
- A1 m
- B2 m
- C7.1e-13 m
- D-1 m
61.Electrons are accelerated through a potential difference of 40 V and diffracted by graphite with atomic spacing 2.1 × 10^-10 m. Calculate the angle of the first minimum of the diffraction pattern, in degrees, to 2 significant figures. Use h = 6.63 × 10^-34 J s, m_e = 9.11 × 10^-31 kg, e = 1.6 × 10^-19 C.
Hard
- A34 °
- B37 °
- C40 °
- D31 °
62.The work function of sodium is 2.28 eV. Calculate the threshold frequency of sodium in Hz. Use h = 6.63 × 10^-34 J s and 1 eV = 1.6 × 10^-19 J. Give your answer to 3 significant figures.
Medium
- A660000000000000 Hz
- B605000000000000 Hz
- C495000000000000 Hz
- D550000000000000 Hz

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