Calculate the wavenumber of photon emitted during transition from the orbit of $n=3$ to $n=2$ in hydrogen…
Calculate the wavenumber of photon emitted during transition from the orbit of $n=3$ to $n=2$ in hydrogen atom. $\left(\mathrm{R}_{\mathrm{H}}=109677 \mathrm{~cm}^{-1}\right)$
$15354.8 \mathrm{~cm}^{-1}$
$82257.8 \mathrm{~cm}^{-1}$
$30515.4 \mathrm{~cm}^{-1}$
$41128.5 \mathrm{~cm}^{-1}$
Solution
The formula for the wave number \((\nu)\) is:
\(\nu=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
where:
- \(R_H=109677 \mathrm{~cm}^{-1}\),
- \(n_1=2\) (lower orbit),
- \(n_2=3\) (higher orbit).
Substitute the values:
\(\begin{gathered}
\nu=109677\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
\nu=109677\left(\frac{1}{4}-\frac{1}{9}\right) \\
\nu=109677\left(\frac{9-4}{36}\right) \\
\nu=\frac{109677 \times 5}{36}=15232.9 \mathrm{~cm}^{-1}
\end{gathered}\)