Explain how the Bohr radius of a hydrogen‑like ion changes when the nuclear charge $Z$ is increased from 1 to 2 or 3. Include the formula and the physical reasoning.

hard 3 marks Class 12 · Physics · Atoms

Answer

The Bohr radius decreases inversely with

Z

; for a hydrogen‑like ion the radius is

a_{0} / Z

, so when

Z

increases the electron is pulled closer to the nucleus.

Step-by-step solution

Step 1: Recall the expression for the Bohr radius of a hydrogen‑like ion: $a_{n} = \frac{n ^{2} a _{0}}{Z}$ where $a_{0} = 5.29 \times 1 0^{- 11} m$ is the Bohr radius of hydrogen.
Step 2: For the ground state $n = 1$ , the radius reduces to $a_{1} = \frac{a _{0}}{Z}$ .
Step 3: When $Z$ increases, the Coulomb attraction $- Z e^{2} / (4 p i v a r e p s i l o n_{0} r)$ becomes stronger, lowering the equilibrium distance where the centripetal force balances the electrostatic attraction.
Step 4: Consequently, the orbital radius shrinks in proportion to $1/ Z$ . For $Z = 2$ the radius is $a_{0} /2$ , and for $Z = 3$ it is $a_{0} /3$ .
Step 5: Physically, a higher nuclear charge pulls the electron closer, making the atom smaller.

Common mistake: Assuming the radius is independent of

Z

or confusing

Z

with

n

Concept tested: Bohr radius dependence on nuclear charge

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked