What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom ?

The correct option is B $6.91\times {10}^{14}Hz$, 434 nm
Since $n_i=5$ and $n_f=2,$ this transition gives rise to a spectral line in the visible region of the Balmer series. From equation.
$\Delta E=2.18\times {10}^{-18}J[\frac{1}{5^2}-\frac{1}{2^2}]$
$=-4.58\times {10}^{-19}J$
Negative sign implies that it is an emission energy.
The frequency of the photon (taking energy in terms of magnitude) is given by
$v=\frac{\Delta E}{h}$
$=\frac{4.58\times {10}^{-19}J}{6.626\times {10}^{-34}Js}$
$\Rightarrow 6.91\times {10}^{14}Hz$
The wavelength of photon emitted $\left(\lambda \right)=\frac{c}{v}$
$=\frac{3.0\times {10}^{8}m{s}^{-1}}{6.91\times {10}^{14}Hz}$
$=4.34\times {10}^{-7}m$ or 434 nm