Question

If the ionization potential of Hydrogen is 13.6 eV, then the ionization potentials of $He^+$ and $Li^{2+}$ are respectively

• A. 27.2 eV and 40.8 eV
• B. 54.4 eV and 122.4 eV
• C. 4.35 x $10^{-18}$ J & 6.53 x $10^{-18}$ J
• D. 8.7 x $10^{-18}$ J & 1.96 x $10^{-17}$ J

Answer 1

Answer: (B)

54.4 eV and 122.4 eV

Answer 2

The ionization potential (IP) for a hydrogen-like species (an atom or ion with only one electron) can be calculated using the formula derived from Bohr's model:

$IP = 13.6 \times \frac{Z^2}{n^2} \text{ eV}$

where:

• $Z$ is the atomic number of the element.
• $n$ is the principal quantum number of the electron's initial energy level. For ionization potential, we consider the electron being removed from the ground state, so $n=1$.

Given the ionization potential of Hydrogen ($H$) is 13.6 eV.

For Hydrogen, $Z=1$ and $n=1$.

$IP_H = 13.6 \times \frac{1^2}{1^2} = 13.6 \text{ eV}$

Now, let's calculate the ionization potentials for $He^+$ and $Li^{2+}$:

1. For $He^+$ ion:

$He^+$ is a hydrogen-like species (it has only one electron).

The atomic number for Helium ($He$) is $Z=2$.

The electron is in the ground state, so $n=1$.

$IP_{He^+} = 13.6 \times \frac{Z^2}{n^2} = 13.6 \times \frac{2^2}{1^2} = 13.6 \times 4 = 54.4 \text{ eV}$

2. For $Li^{2+}$ ion:

$Li^{2+}$ is also a hydrogen-like species (it has only one electron).

The atomic number for Lithium ($Li$) is $Z=3$.

The electron is in the ground state, so $n=1$.

$IP_{Li^{2+}} = 13.6 \times \frac{Z^2}{n^2} = 13.6 \times \frac{3^2}{1^2} = 13.6 \times 9 = 122.4 \text{ eV}$

Therefore, the ionization potentials of $He^+$ and $Li^{2+}$ are 54.4 eV and 122.4 eV, respectively.