IGNOU MPH 13 2026 - A mercury vapour lamp emits light of mean wavelength with a

Question

A mercury vapour lamp emits light of mean wavelength $\lambda = 5461 \text{\AA}$ with a spectral spread $\Delta \lambda = 0.012 \text{\AA}$ . Calculate

19 Jan 2026

Answer :

Word Count : 178

To solve this numerically, we first convert the given quantities into meters: [ \lambda = 5461 , \text{Å} = 5461 \times 10^{-10} , \text{m} = 5.461 \times 10^{-7} , \text{m} ] [ \Delta \lambda = 0.012 , \text{Å} = 0.012 \times 10^{-10} , \text{m} = 1.2 \times 10^{-12} , \text{m} ] The corresponding frequency is given by [ \nu = \frac{c}{\lambda} ] where (c = 3 \times 10^8 ,

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