Question

Calculate the wavelength in Angstrom of the photon emitted when an electron in the Bohr orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionization energy of the ground state hydrogen atom is $2.17\times {10}^{-11}$ erg per atom. (Take h= $6.62\times {10}^{-27}erg-s$)

Answer 1

Given,
Electrons in the Bohr orbit returns from $n_2=2$ to $n_1=1$ state.
The ionization energy of the ground state hydrogen atom = $2.17\times {10}^{-11}$ 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, $E_2-E_1=\Delta E\left(\text{when an electron returns from 2 to 1}\right)=\frac{hc}{\lambda }$
$\Rightarrow \lambda =\frac{hc}{E_2-E_1}$ ...(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,
$\Delta E=E_2-E_1=-0.5425\times {10}^{-11}erg-(-2.17\times {10}^{-11})=1.6275\times {10}^{-11}erg$
$\therefore$ from equation (1),
$\lambda =\frac{(6.62\times {10}^{-27})\times (3\times {10}^{10})}{1.6275\times {10}^{-11}}=12.20\times {10}^{-6}cm=1220\stackrel{o}{A}$