Question

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

Answer 1

The energy of an electron in the ground state is given by,

$E_1=-13.6\text{eV}$

When the hydrogen atom is excited by the photon of energy $12.1\text{eV}$

Final energy of the electron will be

$E=-13.6+12.1=-1.51\text{eV}$

The energy of the electron in the $n^{th}$ state of the hydrogen atom is given by,

$E_n=-\frac{13.6}{n^2}\text{eV}$

$\Rightarrow -1.51\text{eV}=-\frac{13.6\text{eV}}{n^2}$

$\Rightarrow n=3$

The formula for number of spectral lines emitted is given by $\frac{n(n-1)}{2}$

$\Rightarrow N=\frac{3(3-1)}{2}=3$

$\ast$ Key Idea : No.of spectral lines emitted will be $=\frac{n(n-1)}{2}$ Where $n=$ orbital electron state.