NEET-XII-Physics
29: Electric Field and Potential
- #19Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.Ans : Given:
Separation between the two charges, r = 0.53 Å = 0.53 × 10-10 m
By Coulomb's Law, force,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{q}_{1}{q}_{2}}{{r}^{2}}``
Here, `` {q}_{1}={q}_{2}=e``
`` \Rightarrow F=\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{{\left(0.53\times {10}^{-10}\right)}^{2}}``
`` =8.2\times {10}^{-8}\,\mathrm{\,N\,}``
Now, mass of an electron, Me = 9.12 × 10-31 kg
The necessary centripetal force is provided by the Coulombian force.
`` \Rightarrow {F}_{e}=\frac{{M}_{e}{v}^{2}}{r}``
`` \Rightarrow {v}^{2}=0.4775\times {10}^{13}``
`` =4.775\times {10}^{12}``
`` \Rightarrow v=2.18\times {10}^{6}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
Page No 121: