Question

15. (a)(i) Explain why X-rays can be used to produce photographs of fractures in bones. (ii) List four uses of x-rays other than in medicine.
(b) State the energy transformation which takes place during the operation of an x-ray tube.
(c)(i) Explain three named dangers to which human beings may be exposed when subjected to large doses of x-rays. (ii) State two precautions that must be taken by persons working with x-rays.
(d) In an x-ray tube, an electron is accelerated from rest towards a tungsten target biased at a potential of 33 kV. Calculate, for the electron, the: (i) kinetic energy; (ii) velocity. [h=6.6×10⁻³⁴Js; Me=9.1×10⁻³¹kg; c=3.0×10⁸ms⁻¹; e=1.6×10⁻¹⁹C]

Accepted Answer

(a)(i) X-rays can be used to photograph bone fractures because bones (being denser, containing calcium) absorb X-rays much more strongly than the surrounding soft tissue, so bones appear as clear shadows/images on X-ray film while soft tissue lets most of the X-rays pass through, allowing fractures (breaks in bone continuity) to be clearly seen.
(ii) Other uses of X-rays: (1) Detecting cracks/flaws in metals (industrial radiography). (2) Security screening (e.g. at airports, for luggage). (3) Studying crystal structures (X-ray crystallography). (4) Detecting fake/forged paintings or artworks.

(b) During the operation of an X-ray tube, electrical energy (used to accelerate the electrons) is converted into kinetic energy of the electrons, which is then converted into X-ray (electromagnetic) energy and heat energy when the fast electrons are suddenly decelerated upon striking the target.

(c)(i) Dangers of large doses of X-rays: (1) Genetic mutations/damage to DNA. (2) Cancer (particularly leukemia) due to cell damage. (3) Skin burns/tissue damage from prolonged exposure.
(ii) Precautions for persons working with X-rays: (1) Wearing lead-lined aprons/protective clothing. (2) Limiting exposure time and maintaining distance from the X-ray source, and using radiation monitoring badges.

(d)(i) Kinetic energy gained = work done by the accelerating potential = eV = 1.6×10⁻¹⁹×33,000 = 5.28×10⁻¹⁵J.
(ii) ½Mev² = KE, so v² = 2×KE/Me = (2×5.28×10⁻¹⁵)/(9.1×10⁻³¹) = 1.16×10¹⁶. v = √(1.16×10¹⁶) ≈ 1.08×10⁸ ms⁻¹.

WAEC Physics 2012 Theory — Question 15

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