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The Irish physicist George Stoney is most famous for introducing the term electron - Leaving Cert Physics - Question 12 - 2022

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The Irish physicist George Stoney is most famous for introducing the term electron. (i) State two properties of the electron. (ii) The photoelectric effect is the ... show full transcript

Worked Solution & Example Answer:The Irish physicist George Stoney is most famous for introducing the term electron - Leaving Cert Physics - Question 12 - 2022

Step 1

State two properties of the electron.

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Answer

The properties of the electron include:

  1. Charge: The electron has a negative charge of approximately -1.6 x 10^-19 coulombs.
  2. Mass: The mass of an electron is approximately 9.11 x 10^-31 kg.

Step 2

Describe an experiment to demonstrate the photoelectric effect.

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Answer

  1. Apparatus: Set up an electroscope, a zinc plate, and a UV lamp.
  2. Method: Shine UV light on the zinc plate, and observe the electroscope. The leaves of the electroscope will diverge due to the release of electrons from the zinc plate.
  3. Observation: Upon exposure to the UV light, the electroscope shows a noticeable charge as electrons are emitted, demonstrating the photoelectric effect.

Step 3

Describe what happens if the frequency of the incident light is below the threshold frequency.

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When the frequency of the incident light is below the threshold frequency, the photoelectric effect does not occur. This means that electrons will not be emitted from the surface of the metal, regardless of the intensity of the light.

Step 4

Calculate the wavelength of light of this frequency.

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Answer

To calculate the wavelength ( λ ) of light, we use the formula:

λ = rac{c}{f}

where:

  • c is the speed of light (approximately 3 x 10^8 m/s)
  • f is the frequency (6.5 x 10^14 Hz)

Substituting the values:

λ = rac{3 x 10^8 ext{ m/s}}{6.5 x 10^{14} ext{ Hz}} = 4.6 x 10^{-7} ext{ m}

Step 5

Calculate the energy of a photon of this frequency.

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The energy (E) of a photon can be calculated using:

E=himesfE = h imes f

where:

  • h is Planck’s constant (approximately 6.63 x 10^-34 J·s)
  • f is the frequency (6.5 x 10^14 Hz)

Substituting the values:

E=(6.63x1034extJs)imes(6.5x1014extHz)4.3x1019extJE = (6.63 x 10^{-34} ext{ J·s}) imes (6.5 x 10^{14} ext{ Hz}) ≈ 4.3 x 10^{-19} ext{ J}

Step 6

How are electrons produced in an X-ray tube?

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Answer

Electrons are produced in an X-ray tube through a process called thermionic emission. This occurs when the cathode is heated, causing electrons to gain enough energy to escape from the surface.

Step 7

How are electrons accelerated in an X-ray tube?

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Electrons are accelerated in an X-ray tube by applying a high voltage between the cathode and the anode. This electric potential difference causes the electrons to accelerate towards the anode.

Step 8

State one property of tungsten that makes it suitable to use as the target.

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Answer

Tungsten has a high melting point and high specific heat capacity, which makes it suitable to withstand the high temperatures generated in an X-ray tube.

Step 9

What material could be used to ensure that the X-rays do not escape from the X-ray tube?

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Answer

Lead is commonly used to shield X-ray tubes as it effectively absorbs and prevents the escape of X-rays, protecting the surrounding environment.

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