MHT CET · Physics · Atomic Physics
In the second orbit of hydrogen atom, the energy of an electron is ' \(E\) '. In the third orbit of helium atom, the energy of the electron will be (atomic number of helium \(=2\) )
- A \(\frac{4 \mathrm{E}}{9}\)
- B \(\frac{4 \mathrm{E}}{3}\)
- C \(\frac{16 \mathrm{E}}{9}\)
- D \(\frac{16 \mathrm{E}}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{16 \mathrm{E}}{9}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & E \propto \frac{Z^2}{n^2} \\ \therefore \quad \frac{E_H}{E_{H e}} & =\frac{\left(Z^2\right)_H}{(n)_H^2} \times \frac{\left(n_{\mathrm{He}}\right)^2}{\left(Z^2\right)_{\mathrm{He}}} \\ & =\frac{1}{2^2} \times \frac{3^2}{2^2}=\frac{9}{16} \\ \therefore \quad E_{H e} & =\frac{16}{9} E_H=\frac{16}{9} E\end{aligned}\)
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