Question

The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level -1.51 eV to -3.4 eV, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

Answer 1

When energy is -1.51 eV then n = 3.
When energy is -3.4 eV then n = 2.
$\therefore \frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{3^2})\frac{1}{\lambda }=1.1\times {10}^7(\frac{1}{4}-\frac{1}{9})=1.1\times {10}^7\times \frac{5}{36}\lambda =\frac{36}{5\times 1.1}\times {10}^{-7}\lambda =6.545\times {10}^{-7}\lambda =6545A^{\circ }$

Thus, it belongs to the Balmer series.