Photons & Wave-Particle Duality (OCR A Level Physics): Flashcards

Exam code: H556

1/35

0Still learning

Know0

  • Define photon.

Cards in this collection (35)

  • Define photon.

    A photon is a massless "packet" or quantum of electromagnetic energy — energy is transferred in discrete amounts rather than continuously.

  • What evidence supports light behaving as a particle?

    The photoelectric effect — light interacts with matter, such as electrons, as a particle.

  • What evidence supports light behaving as a wave?

    The diffraction and interference of light, e.g. in Young's double-slit experiment.

  • In the photoelectric effect, each electron can absorb only a single photon, so only frequencies of light above the .......... frequency will emit a photoelectron.

    In the photoelectric effect, each electron can absorb only a single photon, so only frequencies of light above the threshold frequency will emit a photoelectron.

  • State the two equivalent equations for the energy of a photon.

    E = hf

    E = \frac{hc}{\lambda}

  • True or False?

    The energy of a photon is directly proportional to its wavelength.

    False.

    The energy of a photon is inversely proportional to its wavelength — a longer-wavelength photon has a lower energy than a shorter-wavelength photon.

  • Why did the wave theory of light fail to fully explain the behaviour of light?

    It explained interference and diffraction well, but it could not explain the photoelectric effect.

  • Define electronvolt.

    The electronvolt is the energy gained by an electron travelling, from rest, through a potential difference of one volt.

  • State the value of one electronvolt in joules.

    1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}

  • Why is the electronvolt a useful unit in quantum physics?

    Quantum energies are typically much smaller than one joule, so the electronvolt gives more convenient, manageable values.

  • State the equation relating an electronvolt to the kinetic energy gained by an electron accelerated from rest through a potential difference.

    eV = \frac{1}{2}mv^2

  • How do you convert an energy from eV to J, and from J to eV?

    eV → J: multiply by 1.6 × 10-19

    J → eV: divide by 1.6 × 10-19

  • The electronvolt is derived from the definition of ...........

    The electronvolt is derived from the definition of potential difference.

  • True or False?

    The electronvolt is a unit of potential difference.

    False.

    The electronvolt is a unit of energy, derived from the definition of potential difference.

  • In the experiment to determine Planck's constant using LEDs, what are the independent and dependent variables?

    Independent variable: wavelength of light emitted by the LED (λ)

    Dependent variable: potential difference across the LED (ΔV)

  • State the equation formed by equating the photon energy and the electron energy in the LED experiment.

    e \Delta V = \frac{hc}{\lambda}

  • In the LED experiment, what do you plot to obtain a straight line, and what does the gradient equal?

    Plot potential difference, ΔV, against 1 / wavelength, 1/λ

    Gradient = hc / e

  • How is the Planck constant calculated from the gradient of the LED graph?

    h = \text{gradient} \times \frac{e}{c}

  • The threshold voltage is the minimum p.d. across the LED required before any .......... is able to flow.

    The threshold voltage is the minimum p.d. across the LED required before any current is able to flow.

  • Give one systematic error in the LED experiment and how it can be reduced.

    There is human error in judging the exact voltage at which an LED just begins to glow. This can be reduced by using a black viewing tube in a darkened room, or by plotting current against voltage and extrapolating back to find the threshold voltage.

  • Give one random error in the LED experiment.

    LEDs do not emit a single wavelength — they emit a narrow spectrum around 60 nm wide, so the quoted (central) wavelength introduces uncertainty.

  • True or False?

    It is safe to stare directly at any LED once it just begins to glow.

    False.

    You should never stare directly at the blue LED, even when dimly lit, as it is closest to the UV part of the spectrum.

  • What apparatus is used to demonstrate electron diffraction, and what target does the electron beam pass through?

    An electron diffraction tube. Electrons are accelerated through a high potential (e.g. 5000 V) and directed through a thin film of graphite.

  • What pattern is produced on the fluorescent screen in the electron diffraction experiment, and why?

    A series of concentric rings, because electrons diffract from the gaps between carbon atoms in the graphite, which act like slits.

  • Why does electron diffraction provide evidence for wave-particle duality?

    Diffraction is a wave-like behaviour, yet electrons — normally described as particles — show it, proving that particles can also behave as waves.

  • State the equation relating the kinetic energy of the electrons to the accelerating voltage.

    E_k = \frac{1}{2}mv^2 = eV

  • Graphite is used in this experiment because its .......... structure provides gaps that act as slits for the electron waves.

    Graphite is used in this experiment because its crystalline structure provides gaps that act as slits for the electron waves.

  • What effect does increasing the accelerating voltage have on the electron diffraction rings?

    It reduces the diameter of the rings; a lower accelerating voltage increases the diameter of the rings.

  • True or False?

    If electrons behaved only as particles, a diffraction pattern would still be observed on the screen.

    False.

    If electrons acted purely as particles, they would be distributed uniformly across the screen, and no diffraction pattern would appear.

  • Define de Broglie wavelength.

    The de Broglie wavelength is the wavelength associated with a moving particle, related to its momentum by λ = h/p, demonstrating wave-particle duality for all particles.

  • State the de Broglie equation.

    \lambda = \frac{h}{p}

  • How can the de Broglie wavelength be written in terms of a particle's mass and speed?

    \lambda = \frac{h}{mv} since momentum p = mv.

  • State the de Broglie wavelength of a particle in terms of its kinetic energy.

    \lambda = \frac{h}{\sqrt{2mE}}

  • The de Broglie equation links a particle-like property, .........., to a wave-like property, wavelength.

    The de Broglie equation links a particle-like property, momentum, to a wave-like property, wavelength.

  • True or False?

    The de Broglie equation only applies to electrons.

    False.

    The de Broglie equation applies to all particles, demonstrating wave-particle duality for matter in general.

Sign up to unlock flashcards

or