Question

Ionization potential of hydrogen atom is $ 13.6\, eV $ . Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy $ 12.1\, eV $ . According to Bohr's theory, the spectral lines emitted by hydrogen will be

• A. two
• B. three
• C. four
• D. one

Answer 1

Answer: (B)

Ionization energy corresponding to ionization potential $=-13.6 \,eV$
Photon energy incident $=12.1 \,eV$
So, the energy of electron in excited state
$=-13.6+12.1=-1.5 \,eV$
ie, $ E_{n} =-\frac{13.6}{n^{2}} eV $
$-1.5 =\frac{-13.6}{n^{2}} $
$\Rightarrow n^{2} =\frac{-13.6}{-1.5} \approx 9 $
$\therefore n=3$
ie, energy of electron in excited state corresponds to third orbit
The possible spectral lines are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and
2nd to 1st. Thus, 3 spectral lines are emitted.