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Calculate the wavenumber of photon emitted during transition from the orbit of $n=3$ to $n=2$ in hydrogen atom. $\left(\mathrm{R}_{\mathrm{H}}=109677 \mathrm{~cm}^{-1}\right)$
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The correct answer is:
$15354.8 \mathrm{~cm}^{-1}$
$\bar{v}=109677\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right] \mathrm{cm}^{-1}$
$=109677\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \mathrm{cm}^{-1}$
$=109677\left[\frac{1}{4}-\frac{1}{9}\right] \mathrm{cm}^{-1}$
$=109677\left[\frac{9-4}{9 \times 4}\right] \mathrm{cm}^{-1}$
$=109677\left[\frac{5}{36}\right] \mathrm{cm}^{-1}$
$=15232.9 \mathrm{~cm}^{-1}$
$=109677\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \mathrm{cm}^{-1}$
$=109677\left[\frac{1}{4}-\frac{1}{9}\right] \mathrm{cm}^{-1}$
$=109677\left[\frac{9-4}{9 \times 4}\right] \mathrm{cm}^{-1}$
$=109677\left[\frac{5}{36}\right] \mathrm{cm}^{-1}$
$=15232.9 \mathrm{~cm}^{-1}$
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