How Niels Bohr found radius of hydrogen-like atom

According to Bohr, electrons move in stationary orbits where they do not radiate energy, hence energy of an orbit remains constant with time.

Bohr assumed stationary orbits (not proved mathematically) This explains atomic stability Energy is quantized and constant in each orbit.

Electrostatic force = Centripetal force

Step 1: Electron is attracted to nucleus, so:mv2r=14πε0Ze2r2\frac{mv^2}{r} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Z e^2}{r^2}

Step 2 : Quantization of angular momentum

mvr=nh2πmvr = \frac{nh}{2\pi}


Step 3: Substitute v into Step 1

m(nh2πmr)2=14πε0Ze2rm \left( \frac{nh}{2\pi mr} \right)^2 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Z e^2}{r}


Step 4: Simplify

n2h24π2mr2=14πε0Ze2r\frac{n^2 h^2}{4\pi^2 m r^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Z e^2}{r}


Step 5: Solve for radius r

r=4πε0n2h24π2mZe2r = \frac{4\pi \varepsilon_0 n^2 h^2}{4\pi^2 m Z e^2}

Simplify:r=ε0h2πme2n2Zr = \frac{\varepsilon_0 h^2}{\pi m e^2} \cdot \frac{n^2}{Z}


Final Result

rn=n2a0Zr_n = \frac{n^2 a_0}{Z}

where,a0=ε0h2πme2=0.529A˚a_0 = \frac{\varepsilon_0 h^2}{\pi m e^2} = 0.529 \, \text{Å}

Key Results

  • rnn2r_n \propto n^2
  • rn1Zr_n \propto \frac{1}{Z}
  • Valid for hydrogen-like species (H, He⁺, Li²⁺)

Bohr model, hydrogen-like atom, radius derivation, angular momentum quantization, electrostatic force

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