The ionization energy for a hydrogen atom in the n = 2 state is 328 kJ·mol⁻¹.
The <em>first ionization energy</em> of hydrogen is 1312.0 kJ·mol⁻¹.
Thus, H atoms in the <em>n</em> = 1 state have an energy of -1312.0 kJ·mol⁻¹ and an energy of 0 when <em>n</em> = ∞.
According to Bohr, Eₙ = k/<em>n</em>².
If <em>n</em> = 1, E₁= k/1² = k = -1312.0 kJ·mol⁻¹.
If <em>n</em> = 2, E₂ = k/2² = k/4 = (-1312.0 kJ·mol⁻¹)/4 = -328 kJ·mol⁻¹
∴ The ionization energy from <em>n</em> = 2 is 328 kJ·mol⁻¹
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Answer (AR)
Explanation: mn has an atomic number of 25 . Simplify use this information to obtain its electronic configuration . MN 1s2 2s2 2p6 3s2 3p6 4s2 3d5
The answer to this question is: TRUE
