KCET · Chemistry · Structure of Atom
The energy of electron in the \( n^{\text {th }} \) Bohr orbit of H-atom is
- A \( \frac{-13.6}{n^{2}} \mathrm{eV} \)
- B \( \frac{-13.6}{n} \mathrm{eV} \)
- C \( \frac{-13.6}{n^{4}} \mathrm{eV} \)
- D \( \frac{-13.6}{n^{3}} \mathrm{eV} \)
Answer & Solution
Correct Answer
(A) \( \frac{-13.6}{n^{2}} \mathrm{eV} \)
Step-by-step Solution
Detailed explanation
The energy of \( e^{-} \)in the nth Bohr orbit of H-atom is given by expression
\[
\begin{array}{l}
E_{n}=\frac{-2 \Pi^{2} m e^{4} Z^{2}}{n^{2} h^{2}\left(4 \Pi \varepsilon_{0}\right)^{2}} \\
=-13.6 \frac{Z^{2}}{n^{2}} e V
\end{array}
\]
Since \( Z=1 \) for hydrogen above equation can be further simplified to
\[
E_{n}=\frac{-13.6}{n^{2}} \mathrm{eV}
\]
\[
\begin{array}{l}
E_{n}=\frac{-2 \Pi^{2} m e^{4} Z^{2}}{n^{2} h^{2}\left(4 \Pi \varepsilon_{0}\right)^{2}} \\
=-13.6 \frac{Z^{2}}{n^{2}} e V
\end{array}
\]
Since \( Z=1 \) for hydrogen above equation can be further simplified to
\[
E_{n}=\frac{-13.6}{n^{2}} \mathrm{eV}
\]
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