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प्रश्न
In a Coolidge tube, electrons strike the target and stop inside it. Does the target get more and more negatively charged as time passes?
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उत्तर
An electron emitted from the filament undergoes a number of collisions inside the material and loses its kinetic energy before coming to rest. This energy is utilised to give out photons or eject electrons from the atoms of the target. These electrons move to the battery connected to the circuit. Thus, the target does not get more and more negatively charged as time passes.
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संबंधित प्रश्न
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].
(d) 5890 Å - 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].
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X-ray from a Coolidge tube is incident on a thin aluminium foil. The intensity of the X-ray transmitted by the foil is found to be I0. The heating current is increased to increase the temperature of the filament. The intensity of the X-ray transmitted by the foil will be
(a) zero
(b) < I0
(c) I0
(d) > I0
Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristic K or L X-ray. The energy levels of the tungsten atom with an electron knocked out are as follows.
| Cell containing vacancy | K | L | M |
| Energy in keV | 69.5 | 11.3 | 2.3 |
A certain element emits Kα X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.
(Use Planck constant h = 6.63 × 10-34 Js= 4.14 × 10-15 eVs, speed of light c = 3 × 108 m/s.)
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(Use Planck constant h = 6.63 × 10-34 Js= 4.14 × 10-15 eVs, speed of light c = 3 × 108 m/s.)
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