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

2 Answers
Mar 30, 2015

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).

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, #"Z"#.

For one-electron species, the energy of an electron that orbits the nucleus at level #"n"# is given by

#E = -Z^(2)/n^(2) * underbrace(((k_e * e^(2))^(2) * m_e)/(2bar(h)^(2)))_(R_E)#, where

#R_E# represents a constant called the Rydberg energy, #R_E = "13.6 eV"#.

The equation for energy becomes

#E = - R_E * 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 * 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")#