JEE Main 2014 Physics question
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: (a) 700 nm to 1 mm (b) 1 nm to 400 nm (c) nm (d) 1 mm to 0.1 m List II: (i) Vibration of atoms and molecules. (ii) Inner shell electrons in atoms moving from one energy level to a lower level. (iii) Radioactive decay of the nucleus. (iv) Magnetron valve.
Answer: (2)
Solution
700 nm to 1 mm is the infrared region, produced by the vibration of atoms and molecules (i). 1 nm to 400 nm covers the ultraviolet and X-ray region, produced when inner-shell
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