MHT CET · Chemistry · Structure of Atom
Calculate the wavenumber of the photon emitted during transition from the orbit of \(\mathrm{n}=2\) to \(\mathrm{n}=1\) in hydrogen atom. \(\left[R_H=109677 \mathrm{~cm}^{-1}\right]\)
- A \(27419.3 \mathrm{~cm}^{-1}\)
- B \(109677.0 \mathrm{~cm}^{-1}\)
- C \(12064.5 \mathrm{~cm}^{-1}\)
- D \(82257.8 \mathrm{~cm}^{-1}\)
Answer & Solution
Correct Answer
(D) \(82257.8 \mathrm{~cm}^{-1}\)
Step-by-step Solution
Detailed explanation
For hydrogen atom,
\(\bar{v}=109677\left[\frac{1}{{n_{\mathrm{f}}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right] \mathrm{cm}^{-1}\)
Here, \(\mathrm{n}_{\mathrm{i}}=2, \mathrm{n}_{\mathrm{f}}=1\)
\(\begin{aligned}
\therefore \quad \bar{v} & =109677\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \mathrm{cm}^{-1} \\
& =109677\left[\frac{1}{1}-\frac{1}{4}\right] \mathrm{cm}^{-1} \\
& =109677\left[\frac{3}{4}\right] \mathrm{cm}^{-1} \\
& =82257.8 \mathrm{~cm}^{-1}
\end{aligned}\)
\(\bar{v}=109677\left[\frac{1}{{n_{\mathrm{f}}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right] \mathrm{cm}^{-1}\)
Here, \(\mathrm{n}_{\mathrm{i}}=2, \mathrm{n}_{\mathrm{f}}=1\)
\(\begin{aligned}
\therefore \quad \bar{v} & =109677\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \mathrm{cm}^{-1} \\
& =109677\left[\frac{1}{1}-\frac{1}{4}\right] \mathrm{cm}^{-1} \\
& =109677\left[\frac{3}{4}\right] \mathrm{cm}^{-1} \\
& =82257.8 \mathrm{~cm}^{-1}
\end{aligned}\)
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