The wavelength of the light emitted, as an electron jumps from second orbit to first orbit in a hydrogen atom is,

The correct option is A $1.225\times {10}^{-7}\text{m}$
Energy of an electron, in electron-volt, is,
$E_2=-\frac{13.6}{n_2^2},E_1=\frac{-13.6}{n_1^2}$ here $n_2=2\text{for second orbit}$
$n_1=1\text{for first orbit}$

So, $E_1=-\frac{13.6}{1^2}=-13.6\text{eV}$
and, $E_2=-\frac{-13.6}{2^2}=-3.4\text{eV}$

As the electron jumps from second to first orbit,
$\left|\Delta E\right|=|E_1-E_2|=\frac{hc}{\lambda }$
$|-10.2|\times 1.6\times {10}^{-19}=\frac{6.63\times {10}^{34}\times 3\times {10}^8}{\lambda }$

$16.32\times {10}^{-19}=\frac{2\times {10}^{-25}}{\lambda }$

$\therefore \lambda =1225\mathring{\text{A}}$

Hence, $\left(A\right)$ is the correct answer.