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: Formula
Total energy in $n^{th}$ orbit is $E_n \propto \frac{Z^2}{n^2}$.

Step 2: Compare Atoms
For both, the orbit number is the same ($n=2$). * For Hydrogen ($H$): $Z_H = 1$. * For Helium Ion ($He^+$): $Z_{He} = 2$.

Step 3: Calculation
$\frac{E_H}{E_{He}} = \frac{Z_H^2}{Z_{He}^2} = \frac{1^2}{2^2} = \frac{1}{4}$.
Final Answer: (D)