Radius and Energy of Bohr Orbits for Hydrogen-Like Ions
Chemistry · Structure Of Atom · NEET
For a hydrogen-like ion (only one electron, like H, He⁺, Li²⁺), the orbit radius is rₙ = 52.9 × (n²/Z) pm, and the energy is Eₙ = −13.6 × (Z²/n²) eV (or −2.18×10⁻¹⁸ × Z²/n² J). Here n is the orbit number and Z is the atomic number (nuclear charge). Memory hook: "radius LOVES n and HATES Z (n² on top), energy is the OPPOSITE (Z² on top)."
One electron circles a nucleus of charge +Z. Radius grows with n² but shrinks with Z; energy is negative and scales with Z²/n² — note how the two formulas flip n and Z.
Your doubts, answered
What exactly is a "hydrogen-like ion"?
It is any atom or ion that has ONLY ONE electron left. Examples: H (Z=1), He⁺ (Z=2), Li²⁺ (Z=3), Be³⁺ (Z=4). Because they have one electron just like hydrogen, the same Bohr formulas work — you only change the value of Z. That is why NEET loves testing He⁺ and Li²⁺: same formula, bigger Z.
What are the two formulas I must remember?
Radius: rₙ = 52.9 × (n²/Z) pm (52.9 pm = a₀, the Bohr radius). Energy: Eₙ = −13.6 × (Z²/n²) eV, which is the same as −2.18×10⁻¹⁸ × (Z²/n²) J. Notice: in radius, n is on top and Z is on bottom. In energy, Z is on top and n is on bottom. They are opposite, so do not mix them up.
Why does the radius get SMALLER when Z gets bigger?
Bigger Z means a stronger positive charge in the nucleus. A stronger pull drags the electron closer, so the orbit shrinks. Since Z is in the denominator of the radius formula, radius goes down as Z goes up. That is why He⁺ (Z=2) has a smaller orbit than H (Z=1) for the same n.
Why is the energy always negative?
Negative energy means the electron is BOUND (trapped) to the nucleus. Zero energy is when the electron is completely free (removed from the atom). So the more negative the number, the more tightly the electron is held. Bigger Z makes the energy more negative (electron held tighter).
How do I quickly compare two ions or two orbits?
Use ratios instead of full numbers. Radius ∝ n²/Z, Energy ∝ Z²/n². Example: to compare Li²⁺ (Z=3, n=3) with H, just plug into n²/Z. For radius of Li²⁺ n=3: 3²/3 = 3, so r = 52.9 × 3 = 158.7 pm. Ratios save time in the exam.
What does the 52.9 pm and 13.6 eV actually mean?
52.9 pm (or 0.529 Å) is the radius of the FIRST orbit of a normal hydrogen atom (n=1, Z=1). 13.6 eV is the energy (magnitude) of that same first orbit — it is also the ionisation energy of hydrogen. These are just the base values for H; multiply by n²/Z or Z²/n² to get other cases.
How is velocity related here?
Velocity of the electron vₙ = 2.18×10⁶ × (Z/n) m/s. Velocity goes UP with Z and DOWN with n — the opposite pattern to radius. You rarely need the number, but remember v ∝ Z/n for MCQ comparisons.
⚠️ The NEET trap ✗ Using n²/Z for BOTH radius and energy, so a student writes energy of Be³⁺ (Z=4, n=2) as smaller than He⁺ (Z=2, n=1). ✓ Energy uses Z²/n². For He⁺ n=1: Z²/n² = 4/1 = 4 units. For Be³⁺ n=2: Z²/n² = 16/4 = 4 units — they are EQUAL. So if He⁺ ground energy is −x J, Be³⁺ n=2 energy is also −x J. 🧠 Radius formula = n² on top; Energy formula = Z² on top. Flip them and you flip the answer. "Energy Zeroes in on Z²."
Real NEET questions
2022
If the radius of the second Bohr orbit of the He⁺ ion is 105.8 pm, what is the radius of the third Bohr orbit of the Li²⁺ ion?
A · A. 158.7 pm ✓
B · B. 15.87 pm
C · C. 1.587 pm
D · D. 158.7 Å
Solution: Radius rₙ = 52.9 × (n²/Z) pm. First confirm the given data: for He⁺ (Z=2, n=2), r = 52.9 × (4/2) = 52.9 × 2 = 105.8 pm — matches. Now for Li²⁺ (Z=3, n=3): r = 52.9 × (9/3) = 52.9 × 3 = 158.7 pm. Answer is A. (Option D is a trap using Å instead of pm.)
2024
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:
A · A. −x/9
B · B. −4x
C · C. −4x/9
D · D. −x ✓
Solution: Energy Eₙ = −2.18×10⁻¹⁸ × (Z²/n²) J, so Eₙ ∝ Z²/n². For He⁺ ground state (Z=2, n=1): Z²/n² = 4/1 = 4, and this equals −x. For Be³⁺ (Z=4, n=2): Z²/n² = 16/4 = 4. Both give the same factor of 4, so the energy is also −x J. Answer is D. Trap: using n²/Z instead of Z²/n² leads to a wrong option.
2025
Energy and radius of the ground-state Bohr orbit of He⁺ and Li²⁺ are: [Given Rₕ = 2.18×10⁻¹⁸ J, a₀ = 52.9 pm]
A · A. Eₙ(Li²⁺)=−19.62×10⁻¹⁶ J, rₙ(Li²⁺)=17.6 pm; Eₙ(He⁺)=8.72×10⁻¹⁶ J, rₙ(He⁺)=26.4 pm
B · B. Eₙ(Li²⁺)=−8.72×10⁻¹⁶ J, rₙ(Li²⁺)=17.6 pm; Eₙ(He⁺)=−19.62×10⁻¹⁶ J, rₙ(He⁺)=17.6 pm
C · C. Eₙ(Li²⁺)=−19.62×10⁻¹⁸ J, rₙ(Li²⁺)=17.6 pm; Eₙ(He⁺)=−8.72×10⁻¹⁸ J, rₙ(He⁺)=26.4 pm ✓
D · D. Eₙ(Li²⁺)=−8.72×10⁻¹⁸ J, rₙ(Li²⁺)=26.4 pm; Eₙ(He⁺)=−19.62×10⁻¹⁸ J, rₙ(He⁺)=17.6 pm
Solution: Ground state means n=1. Energy Eₙ = −2.18×10⁻¹⁸ × (Z²/n²) J. For Li²⁺ (Z=3): E = −2.18×10⁻¹⁸ × 9 = −19.62×10⁻¹⁸ J. For He⁺ (Z=2): E = −2.18×10⁻¹⁸ × 4 = −8.72×10⁻¹⁸ J. Radius rₙ = 52.9 × (n²/Z) pm. For Li²⁺: r = 52.9/3 = 17.6 pm. For He⁺: r = 52.9/2 = 26.4 pm. All match option C. Note the power is 10⁻¹⁸ (options with 10⁻¹⁶ are traps) and energies are negative (bound electron).
Solved Structure Of Atom NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
