Using Bohr's atomic model, derive the expression for the radius of the $n^{\text{th}}$ orbit of the revolving electron in a hydrogen atom.

$r_n = \dfrac{n^2\varepsilon_0 h^2}{\pi m_e e^2}$.

CBSE Class 12 · 2020 · Modern PhysicsShort Answer

Using Bohr's atomic model, derive the expression for the radius of the

n^{th}

orbit of the revolving electron in a hydrogen atom.

Answer: $r_{n} = \frac{n ^{2} ε _{0} h ^{2}}{π m _{e} e ^{2}}$ .

Coulomb force provides the centripetal force:

\frac{m _{e} v _{n}^{2}}{r _{n}} = \frac{k e ^{2}}{r _{n}^{2}} \Rightarrow m_{e} v_{n}^{2} = \frac{k e ^{2}}{r _{n}} . (1)

Bohr's quantisation:

m_{e} v_{n} r_{n} = \frac{nh}{2 π}

, so

v_{n} = \frac{nh}{2 π m _{e} r _{n}}

(2)

Substituting (2) in (1):

r_{n} = \frac{n ^{2} h ^{2}}{4 π ^{2} k m _{e} e ^{2}} = \frac{n ^{2} ε _{0} h ^{2}}{π m _{e} e ^{2}} (Z = 1) .

r_{n} = \frac{n ^{2} ε _{0} h ^{2}}{π m _{e} e ^{2}} \propto n^{2}

CBSE Class 12 Board · 2020 · Subjective