MHT CET · Physics · Ray Optics
The ratio of the total energy of the \(2^{\text {nd }}\) orbit electron for the hydrogen atom \(\left({ }_1 H^1\right)\) to that of helium ion \(\left(H_e^{+}\right)\left[\left({ }_2^4 \mathrm{He}\right)\right]\) is
- A 4
- B \(\frac{1}{2}\)
- C 2
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Total energy of an electron in a Hydrogen like atom is given by,
\(E=-13.6 \frac{Z^2}{n^2}\)
\(\frac{E_H}{E_{H e}}=\left(\frac{1}{2}\right)^2\left(\frac{2}{2}\right)^2=\frac{1}{4}\)
\(E=-13.6 \frac{Z^2}{n^2}\)
\(\frac{E_H}{E_{H e}}=\left(\frac{1}{2}\right)^2\left(\frac{2}{2}\right)^2=\frac{1}{4}\)
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