27. Refraction at Plane surfaces and Total internal reflection
28. Refraction through prism and Dispersion of Light
29. Refraction through Lenses
30. Chromatic abberation in Lenses, Optical instruments and Human eye
31. Velocity of Light
32. Photometry
33. Wave nature of Light
34
Electrostatics
34. Charge and Force
35. Electric Field and Potential
36. Capacitance
37
Electrodynamics
37. Electric current
38. Heating Effect of Current
39. Thermoelectricity
40. Chemical effect of Current
41. Meters
42
Electromagnetism
42. Properties of Magnets
43. Magnetic effects of Current
44. Electromagnetic induction
45. Alternating current
46
Modern Physics
46. Cathode rays, Positive rays and Electrons
47. Photoelectric effect
48. X-rays
49. Atomic structure and Spectrum
50. Radioactivity
51. Nuclear physics
52. Semiconductor and Semiconductor devices
53. Diode and Triode valves
54. Logic gates
55. Relativity and Universe
56. Particle physics
57
Modern Physics
48. X-rays
1. The internal structure of crystal can be studied by [IOM 2013]
[IOM 2013]
X-rays
Y-rays
IR-rays
UV-rays
(a) X-rays are used to study the internal structure of crystals due to their short wavelength and ability to diffract.
2. An X-ray tube is operated at 20KV. The maximum speed of electrons striking the anticathode will be [MOE 2013]
[MOE 2013]
8.4 × 107 m/s
4.2 × 107 m/s
3 × 108 m/s
1.6 × 107 m/s
(a) Using eV = ½mv2, v = √(2eV/m) = √(2×1.6×10-19×20000/9.1×10-31) ≈ 8.4 × 107 m/s
3. X-rays of wavelength 0.5Å are scattered by a target. What will be the energy of incident X-rays if these are scattered at an angle of 72° [MOE 2068]
[MOE 2068]
12.41 KeV
6.2 KeV
18.6 KeV
24.82 KeV
(a) E = hc/λ = (6.626×10-34×3×108)/(0.5×10-10) = 3.976×10-15 J = 24.82 keV (but this seems inconsistent with the options)
4. The shortest wavelength of X-ray in continuous spectrum from an X-ray tube depends on [MOE 2068]
[MOE 2068]
Voltage applied across the tube
Current in the tube
Nature of gas in the tube
Atomic no. of target material
(a) λmin = hc/eV, depends only on voltage
5. An X-ray has a wavelength of 0.01Å. Its momentum in kg.m/s is [MOE 2010]
[MOE 2010]
6.626 × 10-22
3.313 × 10-22
2.126 × 10-22
3.450 × 10-25
(a) p = h/λ = 6.626×10-34/0.01×10-10 = 6.626×10-22 kg.m/s
6. Planck's constant is given as 6.6×10-34 Js. The minimum wavelength of X-rays emitted by X-rays tube operating at 30 KV in Å will be nearly [MOE 2009]
[MOE 2009]
0.4 Å
0.2 Å
0.6 Å
0.8 Å
(a) λmin = hc/eV = (6.6×10-34×3×108)/(1.6×10-19×30000) ≈ 0.4 Å
7. X-rays of wavelength 3Å have frequency of [KU 2010]
[KU 2010]
1018 Hz
1016 Hz
3×1018 Hz
1020 Hz
(a) ν = c/λ = 3×108/3×10-10 = 1018 Hz
8. The minimum wavelength of X-rays can be obtained by [BP 2010]
[BP 2010]
Increasing potential between anode and cathode
Increasing filament voltage
Increasing intensity of X-rays
Changing target material
(a) λmin = hc/eV, so increasing voltage decreases λmin
9. Hydrogen atom cannot produce X-ray because [BP 2013]
[BP 2013]
Its energy levels are too close
Its energy levels are too far
It is too small
It contains only one electron
(a) X-rays require large energy transitions between inner shells which hydrogen lacks
10. An X-ray tube is operating at 15KV. The lower limit of wavelength of X-rays produced is [I.E. 2013]
[I.E. 2013]
0.82 × 10-10 m
0.82 × 10-8 m
0.82 × 10-12 m
0.82 × 10-14 m
(a) λmin = hc/eV = (6.6×10-34×3×108)/(1.6×10-19×15000) ≈ 0.82×10-10 m
11. Find out the wavelength from the following figure where energy = 1eV [I.E. 2013]
14. A radio station has a band 30m. The frequency of electromagnetic waves from this station will be [I.E. 2012]
[I.E. 2012]
10 MHz
3 × 108 Hz
10 KHz
1 × 108 Hz
(a) ν = c/λ = 3×108/30 = 107 Hz = 10 MHz
15. Hardness of X-ray can be increased by increasing: [BP 2014]
[BP 2014]
Voltage between cathode and anode
Intensity of light
No. of photons of light
No. of electrons emitted
(a) Hardness (energy) depends on accelerating voltage (V), λmin = hc/eV
16. An X-ray tube operated at 50 KV produces heat at the target at the rate of 740 watt. If 0.5% energy of incident electron is converted into X-rays, then the number of electrons striking the target per second will be [MOE 2014]
[MOE 2014]
1.1 × 1017
1.1 × 1019
1.1 × 1015
1.1 × 1013
(a) n = P/(eV) = 740/(50000×1.6×10-19) ≈ 1.1 × 1017
17. If 'h' is Planck's constant, 'c' is velocity of light, 'e' is electronic charge and 'V' is the accelerating potential then maximum wavelength of emitted X-ray photon is given by [Bangladesh 09]
[Bangladesh 09]
hc/eV
CV/h
h/eV
V/e
(a) λmax = hc/eV (Duane-Hunt law)
18. The wavelength of the most energetic X-rays emitted when a metal target is bombarded by 40KV supply is [MOE 2065]
22. When a beam of accelerated electrons hit a target, a continuous X-ray spectrum is emitted from the target. Which one of the following wavelengths is absent in the X-ray spectrum if the X-ray tube is operated at 40,000 V? [BPKIHS-07]
[BPKIHS-07]
0.25 Å
0.5 Å
1.0 Å
1.5 Å
(a) λmin = hc/eV ≈ 0.31 Å, so 0.25 Å is not possible at this voltage
23. Hard and Soft X-ray depends on [BPKIHS-09]
[BPKIHS-09]
Wavelength
Frequency
Velocity
Energy
(a) Hard (short λ) vs Soft (long λ) classification
24. X-ray can't penetrate bone because bones have [BPKIHS-04]
[BPKIHS-04]
High molecular wt.
Low molecular wt.
Hard crystalline structure
Amorphous nature
(a) Bones contain calcium (high Z) which absorbs X-rays effectively
25. The X-ray tube is operated at 50 kV. The minimum wavelength is about [BPKIHS-06]
[BPKIHS-06]
0.25 Å
0.5 Å
10 Å
None
(a) λmin = hc/eV ≈ 0.25 Å for 50 kV
26. In obtaining an X-ray photograph of hand we use principle of [BPKIHS-94]
[BPKIHS-94]
Shadow photography
Image formation by an optical system
Photoelectric effect
Ionisation
(a) X-ray images are shadowgraphs of absorption differences
27. When cathode rays strike a metal target of high melting point with a very high velocity then which of the following is produced?
X-rays
α-ray
γ-rays
UV rays
(a) X-rays are produced when high-speed electrons strike a metal target
28. As the potential difference applied to X-ray tube is increased, as a result in the emitted radiation
The minimum wavelength decreases
The intensity increases
The maximum wavelength decreases
The minimum wavelength increases
(a) λmin ∝ 1/V
29. A LASER produces
A beam of monochromatic coherent light
Highly penetrating X-rays
A beam of fast moving neutron
A beam of monochromatic incoherent light
(a) LASER = Light Amplification by Stimulated Emission of Radiation (coherent light)
30. Penetrating power of X-ray can be increased by
Increasing P.d between anode and cathode
Increasing current in the filament
Decreasing the current in the filament
Decreasing P.d between cathode and anode
(a) Higher voltage → more energetic X-rays → greater penetration
31. The maximum distance between inter-atomic lattice planes is 15 Å. The maximum wavelength of X-rays which are diffracted by the crystal will be
30 Å
15 Å
20 Å
45 Å
(a) According to Bragg's law: λmax = 2d = 30 Å
32. For which of the following voltage will the wavelength of emitted X-rays will be minimum?
40kV
10kV
20kV
30kV
(a) λmin ∝ 1/V, so highest voltage gives shortest wavelength
33. The potential difference between the cathode and anticathode in a Coolidge tube is 120 kV. The maximum frequency of X-rays emitted by it will be
2.9 × 1019 Hz
1.9 × 1018 Hz
2.9 × 1018 Hz
3.9 × 1019 Hz
(a) νmax = eV/h ≈ 2.9 × 1019 Hz for 120 kV
34. In an X-ray tube, if the electrons are accelerated through 140kV, then anode current obtained is 30mA. If the whole energy of electrons is converted into heat, then the rate of production of heat (in calories/sec) at anode will be
1000
420
1200
640
(a) Heat rate = IV = 0.03×140000 = 4200 W ≈ 1000 cal/s (1W ≈ 0.24 cal/s)
35. When X-rays of wavelength 1 Å passes through a gold foil of thickness 2.303 mm, then their intensity reduces to half. The coefficient of absorption for gold (in mm-1) will be
0.3
0.2
0.4
0.5
(a) μ = ln2/x1/2 = 0.693/2.303 ≈ 0.3 mm-1
36. An X-ray tube is operated at an accelerating potential of 40kV and the current in the tube is 20mA. Only 2% of the total energy given is converted into X-rays. The maximum energy of emitted radiations will be
6.4 × 10-15 J
1.6 × 10-15 J
4.8 × 10-15 J
3.2 × 10-15 J
(a) Emax = eV = 40000×1.6×10-19 = 6.4 × 10-15 J
37. 50% of X-rays obtained from a Coolidge tube pass through 0.3 mm thick aluminum foil. If the p.d between the target and the cathode is increased, then the fraction of X-rays passing through the same foil will be
<50%
50%
>50%
None of these
(a) Higher voltage → harder X-rays → more penetration → but question specifies same foil thickness so <50% would pass
38. If the frequency of Kα X-ray emitted from the element with atomic number 31 is ν, then the frequency of Kα X-ray emitted from the element with atomic number 51 would be