Using Bohr's postulates of the atomic model, derive the expression for the radius of the $n^{\text{th}}$ electron orbit. Hence obtain the expression for Bohr's radius.

$r_n = \dfrac{n^2\varepsilon_0 h^2}{\pi m e^2}$; $a_0 = 0.53\ \text{\AA}$.

CBSE Class 12 · 2014 · Delhi · Modern PhysicsShort Answer

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

n^{th}

electron orbit. Hence obtain the expression for Bohr's radius.

Answer: $r_{n} = \frac{n ^{2} ε _{0} h ^{2}}{π m e ^{2}}$ ; $a_{0} = 0.53 \overset{˚}{A}$ .

Coulomb force provides the centripetal force:

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

Bohr's quantisation of angular momentum:

m v r = \frac{nh}{2 π} \Rightarrow v = \frac{nh}{2 π m r} . (2)

Substituting (2) in (1):

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

For

n = 1

this gives Bohr's radius:

a_{0} = \frac{ε _{0} h ^{2}}{π m e ^{2}} \approx 0.53 \overset{˚}{A}

CBSE Class 12 Board · 2014 · Delhi · Subjective