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NCERT Exemplar · Q30

Q.Wavelengths of different radiations are given below :
λ(A) = 300 nm λ(B) = 300 μm λ(C) = 3 nm λ(D) = 30 Å
Arrange these radiations in the increasing order of their energies.

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The energy of an electromagnetic wave is inversely proportional to its wavelength. Converting all wavelengths to meters and comparing gives the increasing order of energy: B < A < D < C.

The key idea here is the relationship between the energy of a photon and its wavelength. For electromagnetic radiation, each photon carries energy given by E=hcλE = \frac{hc}{\lambda}, where hh is Planck's constant and cc is the speed of light. Since hh and cc are constants, energy is inversely proportional to wavelength — shorter wavelength means higher energy, and longer wavelength means lower energy.

To compare the energies, we first need all wavelengths in the same unit. Let's convert each to meters.

  1. Convert all wavelengths to meters

    • λ(A) = 300 nm = 300×10−9300 \times 10^{-9} m = 3.00×10−73.00 \times 10^{-7} m
    • λ(B) = 300 μm = 300×10−6300 \times 10^{-6} m = 3.00×10−43.00 \times 10^{-4} m
    • λ(C) = 3 nm = 3×10−93 \times 10^{-9} m = 3.00×10−93.00 \times 10^{-9} m
    • λ(D) = 30 Å = 30×10−1030 \times 10^{-10} m = 3.00×10−93.00 \times 10^{-9} m

    Notice that λ(C) and λ(D) are actually equal: 3 nm = 30 Å (since 1 Å = 10−1010^{-10} m and 1 nm = 10−910^{-9} m). So C and D have the same wavelength.

  2. Arrange wavelengths in increasing order

    Smaller wavelength → higher energy. So we sort from smallest to largest:

    • λ(C) = λ(D) = 3.00×10−93.00 \times 10^{-9} m (smallest)
    • λ(A) = 3.00×10−73.00 \times 10^{-7} m
    • λ(B) = 3.00×10−43.00 \times 10^{-4} m (largest)
  3. Translate to increasing order of energy

    Since energy is inversely proportional to wavelength, the increasing order of energy is the reverse of the increasing order of wavelength:

    • Lowest energy: longest wavelength → B
    • Next: A …

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