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Long Answer Questions · Q11

Q.State the postulates of Bohr's atomic model and derive the expression for the energy of an electron in the atom.

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Bohr's three postulates are: (1) electrons revolve in circular orbits, with the centripetal force provided by electrostatic attraction to the nucleus; (2) only orbits with angular momentum mevnrn=nh2πm_ev_nr_n=\frac{nh}{2\pi} are allowed, and electrons in these stable orbits do not radiate; (3) a transition between orbits emits or absorbs a photon whose energy equals the orbit-energy difference. To derive the energy formula, first combine postulates (1) and (2): the centripetal-force condition mevn2rn=14πϵ0Ze2rn2\frac{m_ev_n^2}{r_n}=\frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r_n^2} and the angular-momentum condition mevnrn=nh2πm_ev_nr_n=\frac{nh}{2\pi} are two equations in vnv_n and rnr_n; solving them together gives rn=n2h2ϵ0πmeZe2r_n=\frac{n^2h^2\epsilon_0}{\pi m_eZe^2} and vn=Ze22ϵ0hnv_n=\frac{Ze^2}{2\epsilon_0hn}. The electron's total energy is the sum of kinetic and (negative) potential energy, En=12mevn2−14πϵ0Ze2rnE_n=\frac{1}{2}m_ev_n^2-\frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r_n}; substituting the expressions for vnv_n and rnr_n and simplifying gives En=−meZ2e48ϵ02h2n2E_n=-\frac{m_eZ^2e^4}{8\epsilon_0^2h^2n^2}. Substituting the numerical values of mem_e, e, ϵ0\epsilon_0 and h converts this into the practically useful form En=−13.6Z2n2E_n=-\frac{13.6Z^2}{n^2} eV, valid for hydrogen (Z=1) and any hydrogen-like (single-electron) ion. [!ANSWER] En=−13.6 Z2n2E_n=-\dfrac{13.6\,Z^2}{n^2} eV.

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