Calculate the wavelength in Angstrom of the photon emitted when an electron in the Bohr...

Given,
Returns from the Bohr orbit of electrons $n_2=2to{n}_1=1$ state.
The hydrogen atom's ground state ionization energy $=2.17\times {10}^{-11}erg/atom$
Again, We know
Ionisation energy $=-E_1$
i.e ionisation energy is the energy required to remove an electron from ground state, so indirectly we are provided with the value of $E_1$.
Also, $E_2-E_1=\Delta E$(when an electron returns from 2 to 1)$=\frac{hc}{\lambda }$
$\Rightarrow \lambda =\frac{hc}{E_2-E_1}$ ...(1)
Where, h = Planck's constant
c = speed of light
We know,
$E_n\propto \frac{Z^2}{n^2}$
[Here Z is constant, since we are considering the same atom]
Hence,
$E_n\propto \frac{1}{n^2}$
$\frac{E_1}{E_n}=\frac{n^2}{1}$
We can say that,
$E_n=\frac{E_1}{n^2}$
Hence putting n = 2,
$\Rightarrow E_2=\frac{E_1}{4}=\frac{-2.17\times {10}^{-11}}{4}$ erg/atom$=-0.5425\times {10}^{-11}$ erg/atom
So,
$\Delta E=E_2-E_1=-0.5425\times {10}^{-11}erg-(-2.17\times {10}^{-11})=1.6275\times {10}^{-11}erg$
∴ from equation (1),
$\lambda =\frac{(6.62\times {10}^{-27})\times (3\times {10}^{10})}{1.6275\times {10}^{-11}}=12.20\times {10}^{-6}cm={1220}^{o}A$