What is the wavenumber of the photon emitted during transition from the orbit $\mathrm{n}=5$ to that of $\mathrm{n}=2$ in hydrogen atom? $\left[\mathrm{R}_{\mathrm{H}}=109677 \mathrm{~cm}^{-1}\right]$

The wavenumber for a hydrogen transition is determined by the Rydberg formula: $\bar{\nu} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, where $R_H = 109677\ \mathrm{cm}^{-1}$ and $Z = 1$.

For the $n = 5$ to $n = 2$ transition, $\bar{\nu} = 109677 \left( \frac{1}{4} - \frac{1}{25} \right)$.

Simplifying, $\frac{1}{4} - \frac{1}{25} = \frac{25 - 4}{100} = \frac{21}{100}$.

Thus, $\bar{\nu} = 109677 \times \frac{21}{100} = 23032.17\ \mathrm{cm}^{-1}$.

Comparing with the options, the value corresponds to A $23032\ \mathrm{cm}^{-1}$.

The correct choice is $\boxed{\text{A}}$.