Question

In the ground state of the Hydrogen atom the energy of the electron is E₀ = -13.61 eV. What is the energy of the electron in the ground state of the He⁺ ion?

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What is the energy of the electron in the ground state of the Li⁺⁺ ion?

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The electron in the He⁺ ion is excited to the n = 2 principal state. What is the energy of the electron now?

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What is the energy of the electron in the Li⁺⁺ ion in the n = 2 principal state?

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What is the energy of the electron in the Li⁺⁺ ion in the n = 3 principal state?

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Take element Z = 47 from the periodic table. Ionize it 46 times so that there is only one electron left orbiting around the nucleus. What is the ground state energy of the electron?

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What is the energy of the electron when it is in the n = 8 principal state in the ion above?

Answer 1

(a) The energy of the electron in the ground state of the He⁺ ion is given as ::

E = - (13.61 eV) Z² / n²                                                                                   { eq.1 }

where, Z = atomic number of helium = 2

n = principal state = 1

inserting these values in above eq.

E = - (13.61 eV) (2)² / (1)²

E = - 54.44 eV

(b) The energy of the electron in the ground state of the Li⁺⁺ ion is given as ::

E = - (13.61 eV) Z² / n²         

where, Z = atomic number of lithium = 3

n = principal state = 1

E = - (13.61 eV) (3)² / (1)²

E = - 122.49 eV

(c) The energy of the electron now which is given as ::

For He⁺ ion, n = principal state = 2 , Z = atomis number = 2

then, E = - (13.61 eV) (2)² / (2)²

E = - 13.61 eV

(d) The energy of the electron in the Li⁺⁺ ion in the n = 2 principal state which is given as ::

E = - (13.61 eV) (3)² / (2)²

E = - 30.62 eV

(e) The energy of the electron in the Li⁺⁺ ion in the n = 3 principal state which is given as ::

E = - (13.61 eV) (3)² / (3)²

E = - 13.61 eV

(f) The ground state energy of the electron will be given as ::

For Z = 47 from the periodic table.

n = 1   (principle state)

E = - (13.61 eV) (47)² / (1)²

E = - 30064.4 eV

(g) The energy of the electron, when it is in the n = 8 principal state in the ion above which given as ::

E = - (13.61 eV) (47)² / (8)²

E = - (30064.4 / 64) eV

E = - 469.7 eV