(a) State two; (i) properties of x-rays (ii) reasons to show that x-rays are waves (iii) uses of x-rays other than those in medicine; (iv) hazards of x-rays...

Assessment: WAEC SSCE - Physics - 2004 (Essay) Subject: Physics

Question 1 Report

(a) State two;

(i) properties of x-rays

(ii) reasons to show that x-rays are waves

(iii) uses of x-rays other than those in medicine;

(iv) hazards of x-rays.

 

(b) The potential difference between the cathode and target of an x-ray tube is 5.00 x 10\(^4\)V and the current in the tube is 2.00 x 10\(^{-2}\)A. Given that only one percent of the total energy supplied is emitted as x-radiation, determine the  ospheric pressure.

(i) maximum frequency of the emitted radiation

(ii) rate at which heat is removed from the target in order to keep it at steady temperature. [Planck's constant, h = 6.63 x 10\(^{-34}\) Js, electronic charge e = 1.60 x 10\(^{-19}\) C]

Answer Details

(a)(i) Two properties of X-rays: they travel in straight lines at the speed of light and are not deflected by electric or magnetic fields (uncharged); they penetrate matter and affect photographic plates.

(ii) Two reasons showing X-rays are waves: they undergo diffraction (and interference) when passed through crystals; they travel with the speed of light and are part of the electromagnetic spectrum (they are not deflected by fields).

(iii) Two uses of X-rays other than in medicine: detecting flaws/cracks in metal castings and welds (industrial radiography); studying crystal structure by X-ray diffraction (crystallography); security screening of luggage.

(iv) Two hazards of X-rays: they damage/destroy living body cells and tissues; prolonged exposure can cause burns, cancer or genetic mutation.

(b) \(V = 5.00 \times 10^{4}\ \text{V}\), \(I = 2.00 \times 10^{-2}\ \text{A}\).

(i) Maximum frequency: the fastest electrons give up all their energy \(eV\) to a single photon, so \(hf_{max} = eV\):

\[ f_{max} = \frac{eV}{h} = \frac{1.60 \times 10^{-19} \times 5.00 \times 10^{4}}{6.63 \times 10^{-34}} \] \[ f_{max} = \frac{8.0 \times 10^{-15}}{6.63 \times 10^{-34}} = 1.21 \times 10^{19}\ \text{Hz} \]

(ii) Rate of heat removal: total power supplied is:

\[ P = VI = 5.00 \times 10^{4} \times 2.00 \times 10^{-2} = 1000\ \text{W} \]

Only 1% becomes X-radiation, so 99% appears as heat that must be removed:

\[ P_{heat} = 0.99 \times 1000 = 990\ \text{W} \]

Heat is removed at the rate of \(990\ \text{J s}^{-1}\) (990 W).

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