Question

The ionization potential of a 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

The correct option is B three

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 line are when electron jumps from orbit 3rd to 2nd; 3rd to 1st and 2nd to 1st. Thus, 3 spectral lines are emitted.