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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. A... 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

An X-ray tube comprises three primary components: the cathode, anode, and a vacuum tube. The electrons are produced at the cathode through the process of thermionic emission, where heat causes the cathode to emit electrons. These electrons are then accelerated towards the anode due to a high voltage applied across the tube.

When the high-speed electrons collide with the anode (target), they suddenly decelerate. This rapid deceleration causes the production of X-ray photons, which emerge from the tube for diagnostic imaging.

Labelled Diagram of an X-ray Tube

Step 2

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

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Answer

To determine the maximum velocity of an electron, we can use the equation derived from the energy gained by the electron when accelerated through a potential difference (V): K.E.=eV=12mv2K.E. = eV = \frac{1}{2}mv^2

Rearranging gives: v=2eVmv = \sqrt{\frac{2eV}{m}} Where: e = charge of the electron = 1.6×1019C1.6 \times 10^{-19} C V = 50,000 V (from the question) m = mass of an electron = 9.11×1031kg9.11 \times 10^{-31} kg

Thus, v=2×(1.6×1019)×500009.11×1031=1.3×108m/sv = \sqrt{\frac{2 \times (1.6 \times 10^{-19}) \times 50000}{9.11 \times 10^{-31}}} = 1.3 \times 10^8 m/s

Step 3

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

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Answer

To find the minimum wavelength (λ\lambda) of the X-rays, we use the equation relating energy and wavelength: E=hcλE = \frac{hc}{\lambda} Where: h = Planck’s constant = 6.63×1034Js6.63 \times 10^{-34} J \, s c = speed of light = 3.0×108m/s3.0 \times 10^8 m/s

The energy of the X-ray photons produced corresponds to the electric potential energy: E=eV=1.6×1019C×50000V=8.0×1015JE = eV = 1.6 \times 10^{-19} C \times 50000 V = 8.0 \times 10^{-15} J

Now substituting into the wavelength equation: λ=hcE=(6.63×1034)(3.0×108)8.0×1015=2.5×1011m\lambda = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{8.0 \times 10^{-15}} = 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 when it absorbs electromagnetic radiation of sufficient frequency. The emitted electrons are known as photoelectrons. This effect is significant because it demonstrates the particle-like behavior of light, supporting the theory that light consists of photons.

Step 5

Describe a laboratory experiment to demonstrate the photoelectric effect.

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Answer

Apparatus:

  • Gold leaf electroscope

Procedure:

  1. Charge the electroscope negatively using a rod.
  2. Shine ultraviolet light onto the metal plate of the electroscope.
  3. Observe that the leaves of the electroscope collapse after exposure to UV light.

Result: The collapse of the leaves indicates that electrons are being emitted from the metal due to the absorption of UV light, demonstrating the photoelectric effect.

Step 6

Outline Einstein’s explanation of the photoelectric effect.

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Answer

Albert Einstein explained the photoelectric effect by proposing that light consists of packets of energy called photons. He indicated the following key points:

  1. Each photon carries a discrete amount of energy, given by the equation: E=hfE = hf where hh is Planck's constant and ff is the frequency of the light.
  2. A photon must have sufficient energy to overcome the work function of the material to liberate an electron.
  3. If the energy of the incoming photon is greater than the work function, the excess energy is converted into kinetic energy of the emitted electron.

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