Respuesta :
The ionization energy for a hydrogen atom in the n = 2 state is 328 kJ·mol⁻¹.
The first ionization energy of hydrogen is 1312.0 kJ·mol⁻¹.
Thus, H atoms in the n = 1 state have an energy of -1312.0 kJ·mol⁻¹ and an energy of 0 when n = ∞.
According to Bohr, Eₙ = k/n².
If n = 1, E₁= k/1² = k = -1312.0 kJ·mol⁻¹.
If n = 2, E₂ = k/2² = k/4 = (-1312.0 kJ·mol⁻¹)/4 = -328 kJ·mol⁻¹
∴ The ionization energy from n = 2 is 328 kJ·mol⁻¹ .
The ionization energy in Kj/mol, for hydrogen atoms initially in the n=2 energy level is;
-5.45 × 10^(-22) J/mol.
According to Bohr's atomic model, the ionization energy of an hydrogen atom is related to the energy level of the atoms as follows;
E = k/n²
where E = ionization energy
k = first ionization energy of hydrogen =
-2.18 × 10^(-18).
and n = energy level = 2
Therefore, the ionization energy of hydrogen atoms in the energy.
E = {-2.18 × 10^(-18)}/2².
E = -5.45 × 10^(-19) J/mol
E = -5.45 × 10^(-22) J/mol.
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