# The energy of a one-electron atom or ion is given by Bohr’s equation. Which one of the following one-electron species has the lowest energy ground state? a)Li2+ b) He+ c)Be3+ d) B4+ e) H

Mar 30, 2015

The correct answer is d) B⁴⁺ has the lowest energy ground state.

Bohr's equation is

${E}_{n} = - \frac{{Z}^{2} R}{n} ^ 2$

For the ground state, $n = 1$, so

${E}_{1} = - {Z}^{2} R$

If $Z = 1$, $E = - R = \text{-13.6 eV}$

Since $R = \text{13.6 eV}$, the formula becomes

${E}_{1} = - 13.6 {Z}^{2} \text{eV}$

We can see that as $Z$ increases, ${E}_{1}$ decreases, but let's do the calculations anyway.

• For $\text{H}$: $Z = 1$; ${E}_{1} = \text{-13.6 eV" × 1^2 = "-13.6 eV}$
• For ${\text{He}}^{+}$: $Z = 2$; ${E}_{1} = \text{-13.6 eV" × 2^2 = "-54.4 eV}$
• For ${\text{Li}}^{2 +}$: $Z = 3$; ${E}_{1} = \text{-13.6 eV" × 3^2 = "-122.4 eV}$
• For ${\text{Be}}^{3 +}$: $Z = 4$; ${E}_{1} = \text{-13.6 eV" × 4^2 = "-218 eV}$
• For ${\text{B}}^{4 +}$: $Z = 5$; ${E}_{1} = \text{-13.6 eV" × 5^2 = "-340 eV}$

The ground state energy is lowest for $Z = 5$ (boron).

Mar 30, 2015

The answer is d) ${B}^{4 +}$.

The lowest energy ground state will belong to the atom or ion that has the largest atomic number, $\text{Z}$.

For one-electron species, the energy of an electron that orbits the nucleus at level $\text{n}$ is given by

$E = - {Z}^{2} / {n}^{2} \cdot {\underbrace{\frac{{\left({k}_{e} \cdot {e}^{2}\right)}^{2} \cdot {m}_{e}}{2 {\overline{h}}^{2}}}}_{{R}_{E}}$, where

${R}_{E}$ represents a constant called the Rydberg energy, ${R}_{E} = \text{13.6 eV}$.

The equation for energy becomes

$E = - {R}_{E} \cdot {Z}^{2} / {n}^{2}$, where

$Z$ - the atomic number;
$n$ - the energy level;

Since you're interested in ground state energy, $n = 1$ and so

$E = - {R}_{E} \cdot {Z}^{2}$

Of the species you've listed, the highest atomic number belongs to boron, $B$, for which $Z = 5$. This means that the ${B}^{4 +}$ ion will have the lowest ground state energy

E = -"13.6 eV" * 5^(2) = color(green)(-"340 eV")