Question 411: Atoms & Nuclei

Question

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

Accepted Answer

Step 1: Recall energy formula for hydrogen-like species
For a hydrogen-like atom/ion (nuclear charge $Z$):
$ E_n = - \frac{13.6\, Z^2}{n^2}\ 	ext{eV} $
Step 2: Energy of electron in H atom ($Z=1$, $n=2$)
$ E_{	ext{H}, n=2} = -\frac{13.6 (1)^2}{2^2} = -\frac{13.6}{4} = -3.4\ 	ext{eV} $
Step 3: Energy of electron in He$^+$ ion ($Z=2$, $n=2$)
$ E_{	ext{He}^+, n=2} = -\frac{13.6 (2)^2}{2^2} = -\frac{13.6 	imes 4}{4} = -13.6\ 	ext{eV} $
Step 4: Ratio
We are asked for the ratio of the
total energies
(taking the values as they are, not just magnitudes unless specified).
$ \frac{E_{	ext{H}, n=2}}{E_{	ext{He}^+, n=2}} = \frac{-3.4}{-13.6}  = \frac{1}{4} $
Ã¢ÂÂ Final Answer:
Option D: $	frac{1}{4}$

Answer

4

Answer

2

Answer

$\frac{1}{2}$

Answer

$\frac{1}{4}$

Atoms & Nuclei

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

Choose the correct answer: