Physics · Atoms · NEET
Yes. The Bohr model works for any hydrogen-like (hydrogenic) species, meaning a nucleus of charge +Ze with only ONE electron. He+ (Z=2) and Li2+ (Z=3) both have a single electron, so the model applies exactly like it does for hydrogen. You just replace the charge e in the nucleus with Ze. NCERT states this directly: hydrogenic atoms are hydrogen, singly ionised helium, doubly ionised lithium, and so on.
Both come from the force balance. The Coulomb pull is proportional to Z (nucleus charge Ze times electron charge e). In the radius formula r is proportional to 1/Z, so radius drops linearly with Z. Energy depends on BOTH the charge product (one Z) AND the smaller radius (another 1/r brings a second Z). Multiplying gives Z x Z = Z². So energy goes as Z² while radius goes as 1/Z.
Nothing in the shape of the formulas, only the value of Z. Hydrogen has Z=1, so the familiar numbers appear: r1 = 0.53 Angstrom, E1 = -13.6 eV. For He+ (Z=2) and Li2+ (Z=3) the SAME formulas apply but with Z inserted. This is why He+ is smaller and needs far more energy to ionise than hydrogen.
Ionisation energy is the energy to remove the electron from the ground state (n=1) to infinity. Ground state energy E1 = -13.6 (Z²/n²) = -13.6 x (2²/1²) = -54.4 eV for He+. Ionisation energy is the magnitude of this, so it equals 54.4 eV. That is 4 times hydrogen's 13.6 eV, because energy scales as Z² and 2² = 4.
A neutral helium atom has TWO electrons. Those two electrons repel each other, and the Bohr model has no way to handle electron-electron repulsion, so it fails. He+ has lost one electron, leaving a single electron around the nucleus. With only one electron there is no electron-electron interaction, so the model works perfectly. The rule is: one electron equals Bohr-valid; two or more equals Bohr fails.
Use r = 0.53 (n²/Z) Angstrom with Z=3, n=1. r = 0.53 x (1/3) = 0.177 Angstrom, about 0.18 Angstrom. It is one-third the hydrogen radius because a triple nuclear charge pulls the single electron three times closer.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Ions that have a nucleus of charge +Ze but only a single orbiting electron. Examples given in NCERT are hydrogen (Z=1), singly ionised helium He+ (Z=2), and doubly ionised lithium Li2+ (Z=3). Because there is only one electron, there is no electron-electron repulsion, so the Bohr model applies exactly.
Radius r = 0.53 (n²/Z) Angstrom (shrinks as 1/Z). Velocity v = 2.18x10^6 (Z/n) m/s (grows with Z). Energy E = -13.6 (Z²/n²) eV (grows as Z²). Angular momentum mvr = n h/2pi is unchanged, it never depends on Z.
E1 = -13.6 (Z²/n²) = -13.6 x (3²/1²) = -13.6 x 9 = -122.4 eV. So the ionisation energy of Li2+ is 122.4 eV, nine times that of hydrogen, because 3² = 9.
No. Bohr's quantisation condition mvr = n h/2pi has no Z in it. Even though velocity increases with Z and radius decreases with Z, their product times mass stays quantised at n h/2pi. The two Z-effects cancel in the angular momentum.
The wave number formula becomes 1/lambda = R Z² (1/n1² - 1/n2²). Since Z²=4 for He+, every spectral line has 4 times the wave number (one-fourth the wavelength) of the matching hydrogen line. This is why some He+ lines overlap the hydrogen series region.
