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Wize University Physics Textbook (Master) > Modern Physics

Bohr's Model of the Atom

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Bohr’s Model of the Atom




The Hydrogen atom is the simplest atom described by quantum theory. The electrons orbit at specific distances from the nucleus, and they can only have specific energy values: they are quantized.

The orbital radius is given by:

 rn=a0 n2 \boxed{ \ r_n=a_0\ n^2 \ }


  • nn is the principal quantum number and represents the number of the energy level
  • a0a_0 is the Bohr Radius, and its value is a0=5.29×10−11 ma_0=5.29\times10^{-11} \ m



The energy levels are given by:

 En=−13.6n2 eV \boxed{ \ E_n=-\dfrac{13.6}{n^2}\ \text{eV} \ }

with n=1,2,3,4,....n=1,2,3,4,....

and n=1n=1 is the ground state.



An electron may transition between levels by either emitting or absorbing a photon of the exact energy corresponding to the gap between the levels:

 hcλ=13.6 eV(1nf2−1ni2) \boxed{ \ \dfrac{hc}{\lambda}=13.6\ \text{eV}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \ }


where nfn_f is the final shell and nin_i is the initial shell.


Alternately, we have:

 1λ=R(1nf2−1ni2) \boxed{ \ \dfrac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \ }


where R=13.6 eVhc=1.097×107  m−1R=\dfrac{13.6\ eV}{hc}=1.097\times 10^7 \ \ m^{-1} is the Rydberg constant.



Wize Concept
  • When the electron moves to a higher energy level, it needs to absorb the photon.
  • When the electron moves to a lower energy level, it will emit the photon.


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Balmer Series
  • Transitions to n=2\bcth {n=2} from the higher shells. They correspond to light in the visible range.






Lyman Series

  • Transitions to n=1\bcth {n=1} (the ground state). They correspond to light in the ultraviolet range.

Example: Energy Levels


a) Calculate the energy required in the electron transition from the n=3n = 3 excited state to the n=2n = 2 excited state in the Hydrogen atom. Is the photon emitted or absorbed?

b) Calculate the wavelength of this photon.


Part a)


The energy required is given by the difference:

ΔE=13.6(1nf2−1ni2)\Delta E=13.6\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)

=13.6(122−132)=13.6\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)

=1.89=1.89 (eV)
=1.89×1.602×10−19=1.89\times1.602\times10^{-19} (J)

=3.03×10−19=3.03\times10^{-19} (J)

Since the electron falls to a lower energy level, the photon is emitted.


Part b)


The wavelength is given by:

E=hcλ   →   λ=hcEE=\dfrac{hc}{\lambda} \ \ \ \to \ \ \ \lambda=\dfrac{hc}{E}

=(6.626×10−34)(3×108 )3.03×10−19=\dfrac{(6.626\times10^{-34})(3\times10^8\ )}{3.03\times10^{-19}}

=6.57×10−7=6.57\times10^{-7} (m)

=657=657 (nm)

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Practice: Transitions


Visible light ranges from 390390 to 700700 nm. Determine three electron shell transitions where a photon in the visible light range is emitted.