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Electromagnetic Waves question

2016 · Shift 0 · Q42
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Electromagnetic Waves question

2016 · Shift 0 · Q42

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
Arrange the following electromagnetic radiations per quantum in the order of increasing energy : A : Blue light B : Yellow light C : X-ray D : Radiowave.
  1. A
    C, A, B, D
  2. B
    B, A, D, C
  3. C
    D, B, A, C
  4. D
    A, B, D, C
View written solutionFree

Correct answer: C

  1. The energy of one quantum (one photon) of electromagnetic radiation is given by

E=hν=hcλE = h\nu = \frac{hc}{\lambda}E=hν=λhc​

where:

  • hhh = Planck's constant
  • ν\nuν = frequency
  • λ\lambdaλ = wavelength

So, higher frequency means higher energy, and equivalently, smaller wavelength means higher energy.

  1. Now compare the given radiations:
  • Radiowave (D): very low frequency, so lowest energy
  • Yellow light (B): visible light, lower frequency than blue light
  • Blue light (A): visible light, higher frequency than yellow light
  • X-ray (C): very high frequency, so highest energy

Thus, the order of increasing energy is:

D<B<A<CD < B < A < CD<B<A<C

  1. Match with the options:
  • A: C,A,B,DC, A, B, DC,A,B,D ❌ decreasing mostly
  • B: B,A,D,CB, A, D, CB,A,D,C ❌ radiowave should come first
  • C: D,B,A,CD, B, A, CD,B,A,C ✅
  • D: A,B,D,CA, B, D, CA,B,D,C ❌ blue has more energy than yellow, not less

Therefore, the correct option is:

C: D,B,A,C\boxed{\text{C: } D, B, A, C}C: D,B,A,C​

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