Question

The ratio of the total energy of the (2^{nd}) orbit electron for the hydrogen atom ((^{1}H)) to that of a helium ion ((He^{+})) is :

• A. (4)
• B. (2)
• C. (\frac{1}{2})
• D. [suspicious link removed]

Hint:

Energy becomes more negative (lower) as the atomic number $Z$ increases.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question compares the energy of electrons in the same orbit (\(n=2\)) for two different single-electron species: Hydrogen and Helium ion.
Step 2: Key Formula or Approach:
The total energy of an electron in the \(n^{\text{th}}\) Bohr orbit is:
\[ E_{n} = -13.6 \frac{Z^{2}}{n^{2}} \text{ eV} \]
where \(Z\) is the atomic number.
Step 3: Detailed Explanation:
For both cases, the orbit number is \(n = 2\).
Therefore, \(E \propto Z^{2}\).
For Hydrogen (\(^{1}\text{H}\)), \(Z_{H} = 1\).
For Helium ion (\(\text{He}^{+}\)), \(Z_{\text{He}} = 2\).
The ratio of energies is:
\[ \frac{E_{H}}{E_{\text{He}}} = \frac{Z_{H}^{2}}{Z_{\text{He}}^{2}} = \frac{1^{2}}{2^{2}} = \frac{1}{4} \]
Step 4: Final Answer:
The ratio is \(\frac{1}{4}\).