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JEE Main2014PhysicsElectromagnetic WavesActual

Match List I (Wavelength range of electromagnetic spectrum) with List II (Method of production of these waves) and select the correct option from the options given below the lists. List I List II (a) 700 nm to 1 mm (i) Vibration of atoms and molecules. (b) 1 nm to 400 nm (ii) Inner shell electrons in atoms moving from one energy level to a lower level. (c) < 10 - 3 nm (iii) Radioactive decay of the nucleus. (d) 1

Options

  1. A(a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)
  2. B(a) - (i), (b) - (ii), (c) - (iii), (d) - (iv)
  3. C(a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)
  4. D(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)

Correct answer

B. (a) - (i), (b) - (ii), (c) - (iii), (d) - (iv)

Step-by-step solution

We know, the range of wavelengths between 700   nm  to   1   mm , is referred as Infrared rays. These are produced by heating an object to such an extent that the molecules start vibrating and it radiates infrared waves. When the object is not hot enough such that it can emit visible light, therefore, it radiates Infrared waves. Thus, option (a) in list I matches with Option (i) in list II. The range of wavelengths between 1   nm  to   400   nm , is referred to as ultraviolet

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