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What is the energy level of li2 when it is excited to an energy level of n=5? (k=2.18×10–18 j) select the correct answer below: −2.18×10−18 j −3.27×10−18 j −7.84×10−19 j −11.1×10−18 j

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Final answer:

The energy level of Li2+ when excited to n=5 is −7.84×10−19 J, determined through the Bohr model calculation considering the atomic number of lithium is 3 and the given energy constant.

Step-by-step explanation:

The question is concerned with determining the energy level of Li2 when it is excited to an energy level of n=5 using the energy constant k=2.18×10−18 J. According to the Bohr model, the energy of an electron in a hydrogen-like atom (which Li2+ can be approximated as because it has a single electron) is given by the formula:

E = -×(k×Z²) / n²

Where:

  • E is the energy,
  • k is the energy constant (2.18×10−18 J),
  • Z is the atomic number of the element (Z=3 for lithium),
  • n is the principal quantum number (energy level).

For the Li2+ ion excited to n=5, we would calculate the energy level as follows:

E = -((2.18×10−18 J)×(3²)) / (5²)

E = -((2.18×10−18 J)×(9)) / (25)

E = -((19.62×10−18 J)) / (25)

E = -7.848×10−19 J

Therefore, the energy level of Li2 when it is excited to an energy level of n=5 is −7.84×10−19 J.

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