Question

The energy of the electron in the second and third Bohr orbits of hydrogen atom is $-5.42\times {10}^{-12}erg$ and $-2.41\times {10}^{-12}erg$ respectively. Calculate the wavelength of the emitted radiation when the electron drops from third to second orbit.

Answer 1

$1erg={10}^{-7}J\Rightarrow E_2=5.42\times {10}^{-19}J=3.38eV{E}_3=2.41\times {10}^{-19}J=1.50eV\Delta E_{3\to 2}=3.38-1.5=1.88eVh\times c=1242eV/nm\Delta E_{3\to 2}=\frac{hc}{\lambda }=\frac{1242}{{\lambda }_{\left(nm\right)}}=1.88\Rightarrow \lambda =660.638nm$