MHT CET · Physics · Gravitation
Ionization potential of hydrogen atom is \(13.6 \mathrm{eV}\). Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy \(12.1 \mathrm{eV} .\) According to Bohr's theory, the spectral lines emitted by hydrogen will be
- A two
- B three
- C four
- D one
Answer & Solution
Correct Answer
(B) three
Step-by-step Solution
Detailed explanation
Ionization energy corresponding to ionization potential \(=-13.6 \mathrm{eV}\)
Photon energy incident \(=12.1 \mathrm{eV}\)
So, the energy of electron in excited state
\(
=-13.6+12.1=-1.5 \mathrm{eV}
\)
\(
\begin{array}{lc}
\text { ie, } \quad E_{n}=-\frac{13.6}{n^{2}} \mathrm{eV} \\
-1.5=\frac{-13.6}{n^{2}} \\
\Rightarrow \quad n^{2}=\frac{-13.6}{-1.5} \approx 9 \\
\therefore n=3
\end{array}
\)
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 1 st and 2nd to 1 st. Thus, 3 spectral lines are emitted.
Photon energy incident \(=12.1 \mathrm{eV}\)
So, the energy of electron in excited state
\(
=-13.6+12.1=-1.5 \mathrm{eV}
\)
\(
\begin{array}{lc}
\text { ie, } \quad E_{n}=-\frac{13.6}{n^{2}} \mathrm{eV} \\
-1.5=\frac{-13.6}{n^{2}} \\
\Rightarrow \quad n^{2}=\frac{-13.6}{-1.5} \approx 9 \\
\therefore n=3
\end{array}
\)
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 1 st and 2nd to 1 st. Thus, 3 spectral lines are emitted.
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