## Questions

1.
1. For a given source of X-rays, how would the following be controlled.
1. Intensity
2. The penetrating power
3. The exposure to patients
2. An accelerating potential of 20kv is applied to an X-ray tube.
1. What is the velocity with which the electron strikes the target?
2. State the energy changes that take place at the target.
2. Explain why X-rays are appropriate in study of the crystalline structure materials.
3. Name the metal used to shield X-rays operators from the radiation. Give reasons why it is used.
4. An X-ray tube is operating with an anode potential of 10kV and a current of 15.0 mA.
1. Explain how the
1. Intensity of X-rays from such a tube may be increased.
2. Penetrating power of X- rays from such a tube may be increased
2. Calculate the number of electrons hitting the anode per second.
3. Determine the velocity with which the electrons strike the target.
4. State one industrial use of X-rays.
5. State the properties of X-rays, which makes it possible to detect cracks in bones.
6.  State one difference between hard X-rays and soft X-rays. (1mk)
7. 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:-
1. The K.E of the electrons
2. The frequency of the photons emitted.
8. An X-rays tubes gives photons of 5.9 x 10-15 J of energy. Calculate:-
1. The wavelength of the photons.
2. The accelerating voltage
3. The velocity of the electrons hitting the target.
9. 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)
10. State the purpose of cooling fins in the X-ray tube.
11. 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)
12. State and explain the effect of increasing the EHT in an X- ray tube on the X-rays produced.

1.
1.
1. Heater current or Filament current
2. Anode Potential or operating potential
3.
• 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
2.
1. ½ MeV2 = eV
V= √(2eV)
Me
= √2 x 1.76 x 109 x 20 x 103
= 6.39 x 107 m/s
2. KE- Heat or internal energy and energy of x- rays or radiation.
2. 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.
3. Lead. It has a high atomic mass hence very dense. This enables it to stop or absorb X-rays
4.
1.
1. Increase the filament current
2. Increase the anode potential
2. 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
3. ½ mv2 = ev
V= √2eV
Me
(2 x 1.6 x 10-19 x 10 x 103)½
9.1 x 10-31
= 5.9 x 107 m/s
4. - Detecting flows/ fault in metals or other structures
- Quality control of manufactured items e.g. tyres, thickness of sheets, Paper.
- Analysis of gem stones
5. Highly penetrating in matter
6. Hard X- rays are more penetrating than soft X- rays due to their higher frequency.
7.
1. 1.6 x 10-13 J
2. 2.424 x 1020Hz
8.
1. 3.35 x 10-11m
2. 36,875V
3. 1.3 x 1016 m/s
9.  3.1 x 10-11m.
10. The fins are used to cool the copper rod which conducts heat away from the target when electrons hit the target
11. 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
λ mm
hc = 1.6 x 10-15 Joules
λ mm
λ mm = 6.6 x 10-34 x 3 x 108
1.6 x 10-15
λ mm = 1.24 x 10-10 m
12. Hard x- rays produced higher EHT results in faster electrons hence higher energy x- rays.

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