KCET · Physics · Atomic Physics
Energy of an electron in the second orbit of hydrogen atom is \(E_{2}\). The energy of electron in the third orbit of \(\mathrm{He}^{+}\)will be
- A \(\frac{9}{16} E_{2}\)
- B \(\frac{16}{9} E_{2}\)
- C \(\frac{3}{16} E_{2}\)
- D \(\frac{16}{3} E_{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{16}{9} E_{2}\)
Step-by-step Solution
Detailed explanation
The energy of the electron in the \(n\)th orbit is given by
\(E_{n}=-13.6 \mathrm{eV} \frac{Z^{2}}{n^{2}}\)
\(E_{n} \propto \frac{Z^{2}}{n^{2}}\)
For hydrogen atom, \(Z=1\)
\(E_{n} \propto \frac{1}{n^{2}}\)
In second orbit, \(n=2\)
\(E_{2} \propto \frac{1}{(2)^{2}}\)
In case of \(\mathrm{He}^{+}, Z=2\) and \(n=3\)
\(\begin{aligned}
& E_{3} \propto \frac{(2)^{2}}{(3)^{2}} \\
\therefore \quad & \frac{E_{2}}{E_{3}}=\frac{1}{4} \times \frac{9}{4}=\frac{9}{16} \\
\Rightarrow \quad & E_{3}=\frac{16 E_{2}}{9}
\end{aligned}\)
\(E_{n}=-13.6 \mathrm{eV} \frac{Z^{2}}{n^{2}}\)
\(E_{n} \propto \frac{Z^{2}}{n^{2}}\)
For hydrogen atom, \(Z=1\)
\(E_{n} \propto \frac{1}{n^{2}}\)
In second orbit, \(n=2\)
\(E_{2} \propto \frac{1}{(2)^{2}}\)
In case of \(\mathrm{He}^{+}, Z=2\) and \(n=3\)
\(\begin{aligned}
& E_{3} \propto \frac{(2)^{2}}{(3)^{2}} \\
\therefore \quad & \frac{E_{2}}{E_{3}}=\frac{1}{4} \times \frac{9}{4}=\frac{9}{16} \\
\Rightarrow \quad & E_{3}=\frac{16 E_{2}}{9}
\end{aligned}\)
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