Question

Calculate the wavelength in angstrom of the photon that is emitted within 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 $2 \cdot 17 \times 10^{-11}$ erg per atom.

Answer 1

I. E. $=E_{\infty}-E_{1}$

$\because \quad E_{n}=-\frac{2 \pi^{2} c^{4} m}{n^{2} h^{2}}=-\frac{1}{n^{2}} \times\left(\frac{2 \pi^{2} c^{4} m}{h^{2}}\right)=-\frac{K}{n^{2}}(K \text { is chosen for convenience })$

$\therefore \quad E_{\infty}=-\frac{K}{\infty}=0 ; \quad E_{1}=-K, \quad E_{2}=-\frac{K}{4}$

$\text { Again, I. E. } \quad=2 \cdot 17 \times 10^{-11}=E_{\infty}-E_{1}=-(-K)=K$

$\Delta E=E_{2}-E_{1}=-\frac{K}{4}+K=\frac{3 K}{4}=\frac{3 \times 2 \cdot 17 \times 10^{-11}}{4} erg$

$\because \quad \Delta E=h v=\frac{h c}{\lambda}$

$\therefore \quad \lambda=\frac{h c}{\Delta E}=\frac{6 \cdot 62 \times 10^{-27} \times 3 \cdot 0 \times 10^{10} \times 4}{3 \times 2 \cdot 17 \times 10^{-11}}$

$=12 \cdot 20 \times 10^{-6} cm =1220 Å$