Question

The wavelength in Angstroms of the photon that is emitted when an electron in Bohr orbit n = 2 returns to the orbit n = 1 in the hydrogen atom will be:
The ionization potential of the ground state of hydrogen atom is 2.17 $\times$ 10${}^{-11}$ erg.

• A. 82 nm
• B. 672 nm
• C. 122 nm
• D. none of these

Answer 1

The correct option is C 122 nm

E${}_2$ - E${}_1$ = -2.17 $\times$ 10${}^{-11}$ [$\frac{1}{4}$ - $\frac{1}{1}$] erg [ n${}_1$ = 1, n${}_2$ = 2]
=2.17 $\times$ 10${}^{-11}$ $\times$ $\frac{3}{4}$ erg
E${}_2$ - E${}_1$ = hv = $\frac{c}{\lambda }$h
$\lambda$ = $\frac{hc}{\Delta E}$ = $\frac{6.62\times {10}^{-27}erg\times 3.0\times {10}^{10}cm}{2.17\times {10}^{-11}\times 3/4}$
= $\frac{2.648}{2.17}$ 10${}^{-5}$ cm
=$1.22\times$ 10${}^{-5}$ cm = $1220A^0$