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X - Rays Questions and Answers - Physics Form 4 Topical Revision
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- For a given source of X-rays, how would the following be controlled.
- The penetrating power
- The exposure to patients
- An accelerating potential of 20kv is applied to an X-ray tube.
- What is the velocity with which the electron strikes the target?
- State the energy changes that take place at the target.
- Explain why X-rays are appropriate in study of the crystalline structure materials.
- Name the metal used to shield X-rays operators from the radiation. Give reasons why it is used.
- An X-ray tube is operating with an anode potential of 10kV and a current of 15.0 mA.
- Explain how the
- Intensity of X-rays from such a tube may be increased.
- Penetrating power of X- rays from such a tube may be increased
- Calculate the number of electrons hitting the anode per second.
- Determine the velocity with which the electrons strike the target.
- State one industrial use of X-rays.
- State the properties of X-rays, which makes it possible to detect cracks in bones.
- State one difference between hard X-rays and soft X-rays. (1mk)
- A target was bombarded by electron accelerated by a voltage of 106 V. If all the K.E of the electrons was converted to X-rays, calculate:-
- The K.E of the electrons
- The frequency of the photons emitted.
- An X-rays tubes gives photons of 5.9 x 10-15 J of energy. Calculate:-
- The wavelength of the photons.
- The accelerating voltage
- The velocity of the electrons hitting the target.
- If accelerating voltage in an X-ray tube is 40kV, determine the minimum wavelength of the emitted X-rays. (Electronic charge = -1.6 x 10-19C, planks constant = 6.6 x 10-34 Js, velocity of electromagnetic waves = 3.0 x 108ms-1)
- State the purpose of cooling fins in the X-ray tube.
- X-rays are produced by a tube operating at 1 x 104V. Calculate their wavelength. (Take h= 6.6 x 10-34 Js, e= 1.6 x 10-19 C, c= 3x108ms-1)
- State and explain the effect of increasing the EHT in an X- ray tube on the X-rays produced.
- Heater current or Filament current
- Anode Potential or operating potential
- Covering with protective materials where X- rays are not required
- Minimize exposure time as much as possible
- Reduce no of exposures as much as possible
- ½ MeV2 = eV
= √2 x 1.76 x 109 x 20 x 103
= 6.39 x 107 m/s
- KE- Heat or internal energy and energy of x- rays or radiation.
- X- rays have wavelengths of the order of the lattice spacing; and therefore they can be diffracted; (Diffraction due to short wavelengths of x- rays). In calculation the atomic separation is equal to slit separation- or grating separation. Lead because it is very dense, has high atomic mass.
- Lead. It has a high atomic mass hence very dense. This enables it to stop or absorb X-rays
- Increase the filament current
- Increase the anode potential
- Q= it = 15 x 10-3 A x 1 is = 15 x 10-3C
Electron charge = 1.6 x 10-19 C
No. of electrons in 15 x 10-3C
= 15 x 10-3 = 9.38 x 1016 e/s
1.6 x 10-19
- ½ mv2 = ev
(2 x 1.6 x 10-19 x 10 x 103)½
9.1 x 10-31
= 5.9 x 107 m/s
- - Detecting flows/ fault in metals or other structures
- Quality control of manufactured items e.g. tyres, thickness of sheets, Paper.
- Analysis of gem stones
- Highly penetrating in matter
- Hard X- rays are more penetrating than soft X- rays due to their higher frequency.
- 1.6 x 10-13 J
- 2.424 x 1020Hz
- 3.35 x 10-11m
- 1.3 x 1016 m/s
- 3.1 x 10-11m.
- The fins are used to cool the copper rod which conducts heat away from the target when electrons hit the target
- K.E on input = e.v
= 1.6 x 10-19 C x 1 x 104V
= 1.6 x 10–15 Joules
Energy of x- rays is hf
Where f= c
hc = 1.6 x 10-15 Joules
λ mm = 6.6 x 10-34 x 3 x 108
1.6 x 10-15
λ mm = 1.24 x 10-10 m
- Hard x- rays produced higher EHT results in faster electrons hence higher energy x- rays.
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