Question

Ionization potential of hydrogen atom is $13.6eV$ . Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy $12.1eV$ . 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 $=-13.6eV$
Photon energy incident $=12.1eV$
So, the energy of electron in excited state
$=-13.6+12.1=-1.5eV$
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.