Chemistry · Structure Of Atom · NEET
For a hydrogen-like atom (one electron, nuclear charge Z), En = -RH·Z²/n². Here RH = 2.18 x 10^-18 J per atom (or 13.6 eV, or 1312 kJ/mol). For hydrogen Z=1, so En = -13.6/n² eV. The n is the orbit number (1, 2, 3...). Only the whole numbers are allowed, which is why energy is quantised.
The minus sign means the electron is bound to the nucleus. We fix the zero of energy at n = infinity, where the electron is free and just escaped the atom. Any bound electron has LESS energy than a free one, so its energy is below zero (negative). It does not mean the amount of energy is negative in a physical sense - it is a comparison to the free state. NEET loves testing this sign.
As n increases, n² gets bigger, so RH/n² gets smaller, so En becomes LESS negative (closer to 0). Example: E1 = -13.6 eV, E2 = -3.4 eV, E3 = -1.51 eV. So higher orbits have HIGHER energy (less negative). Students often flip this - remember: closer to nucleus = more negative = more stable.
Z is the atomic number (nuclear charge). For H it is 1, He+ it is 2, Li2+ it is 3, Be3+ it is 4. Energy depends on Z²/n². So He+ in n=1 has energy -13.6 x (2²/1²) = -54.4 eV. NEET often gives you H's energy and asks for a hydrogen-like ion - just scale by Z²/n².
RH (the Rydberg energy) = 2.18 x 10^-18 J/atom = 13.6 eV/atom = 1312 kJ/mol. Use joules when the question gives values in joules, and eV for quick shell energies. 1 eV = 1.6 x 10^-19 J. Note: this energy RH is different from the Rydberg CONSTANT (1.097 x 10^7 m^-1) used in the spectral wavenumber formula - do not mix them up.
Yes. Energy depends only on the ratio Z²/n². For He+ (Z=2, n=1): Z²/n² = 4. For Be3+ (Z=4, n=2): Z²/n² = 16/4 = 4. Same ratio, same energy. NEET 2024 used exactly this trick - the answer was that both energies are equal.
For a Bohr orbit: Total energy E = -RH·Z²/n² (negative). Kinetic energy KE = +RH·Z²/n² = -E (positive). Potential energy PE = -2RH·Z²/n² = 2E (twice as negative). So KE = -E and PE = 2E, giving E = KE + PE = -KE. This is the virial relation and NEET sometimes asks for the ratio.
The energy of an electron in the ground state (n=1) for the He+ ion is -x J. Then the energy of an electron in the n=2 state for the Be3+ ion, in J, is:
Energy and radius of the ground-state Bohr orbit of He+ and Li2+ are: [Given RH = 2.18 x 10^-18 J, a0 = 52.9 pm]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
E1 = -13.6 eV = -2.18 x 10^-18 J = -1312 kJ/mol. This is the ground state and the most stable (most negative) level.
Yes in magnitude. To remove the electron from n=1 (E1 = -13.6 eV) to n=infinity (E = 0) you must add +13.6 eV. So the ionisation energy of hydrogen equals +13.6 eV.
Multiply the hydrogen value by Z². En = -13.6 x Z²/n² eV. For He+ (Z=2) ground state: -13.6 x 4 = -54.4 eV.
RH here is the Rydberg ENERGY = 2.18 x 10^-18 J = 13.6 eV, used for orbit energies. The Rydberg CONSTANT in the wavenumber formula is 1.097 x 10^7 m^-1. They are related but have different units - do not confuse them in NEET.
Because it links energy, radius, ionisation energy and spectral lines. One formula En = -RH·Z²/n² lets NEET test the sign, the Z² scaling and quick numeric ratios - all common single-mark questions in Structure of Atom.