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X-rays have two important uses in medicine: imaging and radiation therapy - Leaving Cert Physics - Question 7 - 2015

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X-rays have two important uses in medicine: imaging and radiation therapy. Describe, with the aid of a labelled diagram of an X-ray tube, how X-rays are produced. ... show full transcript

Worked Solution & Example Answer:X-rays have two important uses in medicine: imaging and radiation therapy - Leaving Cert Physics - Question 7 - 2015

Step 1

Describe, with the aid of a labelled diagram of an X-ray tube, how X-rays are produced.

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Answer

The production of X-rays in an X-ray tube involves several key components: the cathode, anode, and vacuum tube.

  • Diagram: A labelled diagram should include a cathode (negative electrode), an anode (positive electrode), and the vacuum tube where the process takes place.

  • Step 1: Electrons are produced at the cathode through thermionic emission or heating of the cathode. This process causes the cathode to emit electrons due to thermal energy.

  • Step 2: A high voltage is applied between the cathode and anode, causing the emitted electrons to accelerate towards the anode.

  • Step 3: Upon striking the anode, the high-speed electrons collide with the metal target, producing X-rays through interaction with the atoms in the target material.

Step 2

Calculate (i) the maximum velocity of an electron in the tube.

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Answer

To find the maximum velocity ( u) of an electron, we use the formula derived from the kinetic energy gained by the electron:

eV=12mv2eV = \frac{1}{2} mv^2

Where:

  • e = charge of the electron = 1.6×10191.6 \times 10^{-19} C
  • V = potential difference = 50000 V
  • m = mass of electron = 9.11×10319.11 \times 10^{-31} kg

Rearranging gives: v=2eVmv = \sqrt{\frac{2eV}{m}} Substituting in the values: v=2(1.6×1019C)(50000V)9.11×1031kg=1.3×108m/sv = \sqrt{\frac{2(1.6 \times 10^{-19} C)(50000 V)}{9.11 \times 10^{-31} kg}} = 1.3 \times 10^8 \, m/s

Step 3

(ii) the minimum wavelength of the X-rays produced by the tube.

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Answer

The minimum wavelength (\lambda) of X-rays produced can be calculated using the relation:

λ=hceV\lambda = \frac{hc}{eV}

Where:

  • h = Planck's constant = 6.63×1034Js6.63 \times 10^{-34} J \, s
  • c = speed of light = 3.00×108m/s3.00 \times 10^8 \, m/s
  • eV = energy in terms of electron volts = 50000 V

Substituting the values: λ=(6.63×1034Js)(3.00×108m/s)(1.6×1019C)(50000V)=2.5×1011m\lambda = \frac{(6.63 \times 10^{-34} J \, s)(3.00 \times 10^8 \, m/s)}{(1.6 \times 10^{-19} C)(50000 V)} = 2.5 \times 10^{-11} m

Step 4

What is the photoelectric effect?

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Answer

The photoelectric effect refers to the phenomenon where electrons are emitted from a material (typically a metal) when it is exposed to electromagnetic radiation (light) of sufficient frequency.

  • Emission of Electrons: The photons of light must have energy greater than the work function of the material in order to free electrons from the surface.
  • Threshold Frequency: There exists a minimum frequency, known as the threshold frequency, below which no electrons are emitted regardless of the intensity of the light.

Step 5

Describe a laboratory experiment to demonstrate the photoelectric effect.

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Answer

To demonstrate the photoelectric effect, a simple experiment can be conducted using the following apparatus:

  • Apparatus: Gold leaf electroscope, light source (UV lamp or monochromatic light source), and a metal plate.

  • Procedure:

    1. Charge the gold leaf electroscope positively by using an insulating rod and a material that will induce charging.
    2. Bring the metal plate close to the charged electroscope without touching it.
    3. Illuminate the metal plate using the UV lamp.
    4. Observe the behavior of the gold leaves; they should collapse as electrons are emitted from the metal plate and reduce the positive charge on the electroscope.

Step 6

Outline Einstein's explanation of the photoelectric effect.

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Answer

Albert Einstein's explanation of the photoelectric effect states that:

  1. Photons, which are packets of electromagnetic radiation, interact with electrons in the material. Each photon carries a discrete amount of energy given by the equation: E=hfE = hf where h is Planck's constant and f is the frequency of the light.

  2. A photon must give all its energy to an electron to overcome the work function of the material, resulting in electron emission.

  3. If the energy of the photon exceeds the work function (E0E_0), the remaining energy contributes to the kinetic energy of the emitted electron. KE=EE0KE = E - E_0

  4. If the energy is insufficient (below the threshold frequency), no electrons will be emitted regardless of the intensity of the light.

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