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Calculate the wave number and frequency of orange radiation having wavelength \(6300 Å\).
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Verified Answer
The correct answer is:
\(1.587 \times 10^6 \mathrm{~m}^{-1}, 4.761 \times 10^{14} \mathrm{~s}^{-1}\)
For orange radiation (EMR) of \(\lambda=6300 Å\)
(i) Wave number,
\(\bar{v}=\frac{1}{\lambda}=\frac{1}{6300 Å}=\frac{1}{6300 \times 10^{-10} \mathrm{~m}}\)
\(=1.587 \times 10^6 \mathrm{~m}^{-1}\)
(ii) Frequency,
\(\begin{aligned}
\mathrm{v} & =\frac{c}{\lambda}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{6300 \times 10^{-10} \mathrm{~m}} \\
& =4.761 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}\)
(i) Wave number,
\(\bar{v}=\frac{1}{\lambda}=\frac{1}{6300 Å}=\frac{1}{6300 \times 10^{-10} \mathrm{~m}}\)
\(=1.587 \times 10^6 \mathrm{~m}^{-1}\)
(ii) Frequency,
\(\begin{aligned}
\mathrm{v} & =\frac{c}{\lambda}=\frac{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{6300 \times 10^{-10} \mathrm{~m}} \\
& =4.761 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}\)
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