(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 ene...

(a)(i) Why X-rays can photograph bone fractures

X-rays are highly penetrating. They pass easily through soft tissue (flesh) but are strongly absorbed by the denser bone. When X-rays are directed through a limb onto a photographic film or detector, the bones cast dark shadows while the flesh lets the rays through. A break or crack in a bone shows up as a discontinuity (a line or gap) in the bone shadow, so fractures can be seen.

(a)(ii) Four uses of X-rays other than in medicine

• Detecting flaws, cracks and cavities in metal castings and welds (industrial radiography).
• Studying the arrangement of atoms in crystals (X-ray crystallography/diffraction).
• Screening luggage and cargo for hidden or dangerous objects at airports and borders.
• Checking the authenticity of paintings and examining the internal structure of sealed goods.

(b) Energy transformations in an X-ray tube

Electrical energy is converted into the kinetic energy of the fast-moving electrons; when these electrons strike the target, their kinetic energy is converted mainly into heat (about 99%) and a small part into X-ray (electromagnetic) energy.

(c)(i) Three dangers of large doses of X-rays

• They destroy or damage living body cells and tissues, causing radiation burns.
• They can cause cancer, for example leukaemia.
• They can damage reproductive cells, causing genetic mutations or sterility (and can damage the eyes).

(c)(ii) Two precautions

• Shield the body with lead aprons and work behind lead-lined walls/screens.
• Limit the exposure time and dose, and wear a film badge (dosimeter) to monitor the radiation received.

(d)(i) Kinetic energy of the electron

\[ KE = eV = (1.6 \times 10^{-19}) \times (33 \times 10^{3}) = 5.28 \times 10^{-15}\,\text{J} \]

(d)(ii) Velocity of the electron

Using \( \tfrac{1}{2}m v^2 = KE \):

\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2 \times 5.28 \times 10^{-15}}{9.1 \times 10^{-31}}} \]

\[ v = \sqrt{1.16 \times 10^{16}} \approx 1.08 \times 10^{8}\,\text{m s}^{-1} \]

The electron reaches about \( 1.08 \times 10^{8}\,\text{m s}^{-1} \).