JEE Main 2014 — Electromagnetic Waves Question with Solution
From: JEE Main 2014 (Online) 11th April Morning Slot
Question
| List - I | List - II | ||
|---|---|---|---|
| (I) | Doublet of sodium | (A) | Visible radiation |
| (II) | Wavelength corresponding to temperature associated with the isotropic radiation filling all space |
(B) | Microwave |
| (III | Wavelength emitted by atomic hydrogen in interstellar space |
(C) | Short radiowave |
| (IV) | Wavelength of radiation arising from two close energy levels in hydrogen |
(D) | X - rays |
Choose an option
Show full solutionCorrect option: B
Step-by-step explanation
(I) These have a wavelength of 589 nm – 589.6 nm that corresponds to visible light in the yellow region. The correct matching is (I)-(A), Visible radiation. The given wavelength range falls within the visible spectrum.
(II) The temperature is 2.7 K, which corresponds to the emission of microwaves. The correct matching is (II)-(B), Microwaves. The given temperature corresponds to the cosmic microwave background radiation.
(III) The wavelength associated is 21 cm, which corresponds to radiowaves of short wavelength. The correct matching is (III)-(C), Short radiowave. The given wavelength of 21 cm falls within the range of radiowaves.
(IV) This radiation has a frequency of 1,057 MHz, which corresponds to radiowaves of high frequency or short wavelength. The correct matching is (IV)-(B), Microwaves. The given frequency falls within the microwave range.
Therefore, the correct answer is Option B : (I)-(A), (II)-(B), (III)-(C), (IV)-(B).
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This is a previous-year question from JEE Main 2014, covering the Electromagnetic Waves chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.