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

Md Ariful Islam

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Md Ariful Islam

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

Bohr's equation is

$E_n = -(Z^2R)/n^2$

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

$E_1 = -Z^2R$

If $Z = 1$, $E = -R = "-13.6 eV"$

Since $R = "13.6 eV"$, the formula becomes

$E_1 = -13.6Z^2 "eV"$

We can see that as $Z$ increases, $E_1$ decreases, but let's do the calculations anyway.

  • For $"H"$: $Z = 1$; $E_1 = "-13.6 eV" × 1^2 = "-13.6 eV"$
  • For $"He"^+$: $Z = 2$; $E_1 = "-13.6 eV" × 2^2 = "-54.4 eV"$
  • For $"Li"^(2+)$: $Z = 3$; $E_1 = "-13.6 eV" × 3^2 = "-122.4 eV"$
  • For $"Be"^(3+)$: $Z = 4$; $E_1 = "-13.6 eV" × 4^2 = "-218 eV"$
  • For $"B"^(4+)$: $Z = 5$; $E_1 = "-13.6 eV" × 5^2 = "-340 eV"$

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

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