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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
Detailed Solution
Energy of the $n$th level: $E_n = -\dfrac{13.6}{n^2}$ eV. The atom absorbs 12.1 eV and goes from $n = 1$ to the level $n$.
According to the question: $12.1 = 13.6\left(\dfrac{1}{1^2} - \dfrac{1}{n^2}\right)$
$\dfrac{13.6}{n^2} = 13.6 - 12.1 = 1.5$
$\Rightarrow n^2 = \dfrac{13.6}{1.5} \approx 9 \Rightarrow n = 3$
Number of spectral lines emitted when the electron returns from the $n$th level $= \dfrac{n(n - 1)}{2}$
$= \dfrac{(3)(2)}{2} = 3$
The three lines correspond to the transitions $3 \rightarrow 2$, $2 \rightarrow 1$ and $3 \rightarrow 1$.
According to the question: $12.1 = 13.6\left(\dfrac{1}{1^2} - \dfrac{1}{n^2}\right)$
$\dfrac{13.6}{n^2} = 13.6 - 12.1 = 1.5$
$\Rightarrow n^2 = \dfrac{13.6}{1.5} \approx 9 \Rightarrow n = 3$
Number of spectral lines emitted when the electron returns from the $n$th level $= \dfrac{n(n - 1)}{2}$
$= \dfrac{(3)(2)}{2} = 3$
The three lines correspond to the transitions $3 \rightarrow 2$, $2 \rightarrow 1$ and $3 \rightarrow 1$.