-A light beam going north passes a stationary observer. A second observer passes the first observer at a speed of 0.6 c, going south. According to the second observer, the speed of the beam seen by the first observer is
1-0.4 c
2-1.6 c
3-c
4-0.6 c
5-0.8 c
-When light of wavelength λ illuminates the surface of Metal 1, the stopping voltage is V. In terms of V, what will be the stopping voltage if the same wavelength is used to illuminate the surface of Metal 2? The work function of Metal 1 is 5.7 eV and the work function of Metal 2 is 2.1 eV.
1-V – 3.6 volts
2-3.6 volts – V
3-2.1 volts – V
4-V + 3.6 volts
5-5.7 volts – V
-One GeV is equal to
1-1.6×10– 12 J
2-1.6×10– 4 J
3-1.6×10– 19 J
4-1.6×10– 10 J
5-1.6×1012 J
-Consider the following events:
I) A photoelectric effect in which some emitted electrons have kinetic energies greater than the energy of the incident light.
II) A photoelectric effect in which all emitted electrons have kinetic energies less than the energy of the incident light.
III) Compton scattering of monochromatic light from stationary electrons for which the scattered light has a frequency that is greater than that of the incident light.
IV) Compton scattering of monochromatic light from stationary electrons for which the scattered light has a frequency that is less than that of the incident light.
The only possible of the above events are:
Answer 1: C
speed of light is alawasys C irrespect of speed of observer.( this is einstein postulate of relativity).
Answer 2: 4-V + 3.6 volts
Explanation:
Anaswer 3: (4) - 1.6×10– 10 J
Explanation:
1 ev = 1.6*10-19 J
1Gev=109ev
hence 1Gev = 109 * 1.6*10-19 J = 1.6×10– 10 J
Answer 4: III and IV only
Explanation:
in photo electric effect kinetic energy of electron can never exceed the incident energy of radiation but some of (not all) emitted electrons have kinetic energies same as incident energy.
in compton effect, frequency of scattered light may increase or decrease.
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