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Problems · Problem 2.5

Q.Calculate

(a) wavenumber and
(b) frequency of yellow radiation having wavelength 5800 Å.
Rajasthan RbseTextbookSubjective· 2mImportance★★★★★est
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The wavenumber and frequency of electromagnetic radiation are directly linked to its wavelength. For yellow light with λ=5800 A˚\lambda = 5800 \, \text{Å}, the wavenumber is 1.724×106 m−11.724 \times 10^6 \, \text{m}^{-1} and the frequency is 5.17×1014 Hz5.17 \times 10^{14} \, \text{Hz}.

The key here is understanding what wavenumber and frequency actually represent for an electromagnetic wave. Wavenumber tells you how many wavelengths fit into a unit length — it’s the spatial frequency. Frequency tells you how many wave cycles pass a point per second. Both are derived from the wavelength, but they use different constants: wavenumber uses 1/λ1/\lambda, while frequency uses c/λc/\lambda, where cc is the speed of light.

A common mistake is mixing up units. Wavelength is given in angstroms (Å), but standard SI units require metres. So the first step is always conversion.

  1. Convert wavelength to metres. 1 A˚=10−10 m1 \, \text{Å} = 10^{-10} \, \text{m}, so

λ=5800 A˚=5800×10−10 m=5.8×10−7 m.\lambda = 5800 \, \text{Å} = 5800 \times 10^{-10} \, \text{m} = 5.8 \times 10^{-7} \, \text{m}.

  1. Calculate the wavenumber (νˉ\bar{\nu}). Wavenumber is defined as the reciprocal of wavelength in metres:

νˉ=1λ.\bar{\nu} = \frac{1}{\lambda}.

Substituting:

νˉ=15.8×10−7=1.724×106 m−1.\bar{\nu} = \frac{1}{5.8 \times 10^{-7}} = 1.724 \times 10^{6} \, \text{m}^{-1}.

This means about 1.7 million waves of this yellow light fit into every metre.

Tip

Wavenumber is often expressed in cm−1\text{cm}^{-1} in spectroscopy. To get that, convert λ\lambda to cm first: λ=5.8×10−5 cm\lambda = 5.8 \times 10^{-5} \, \text{cm}, so νˉ=1/(5.8×10−5)≈17240 cm−1\bar{\nu} = 1/(5.8 \times 10^{-5}) \approx 17240 \, \text{cm}^{-1}. But the problem asks for SI units, so stick with m−1\text{m}^{-1}. …

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