question_answer

Energy of a hydrogen atom in its ground state is (r = atomic radius of hydrogen atom)            

A) \[-\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}\]

B) \[\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}\]

C) \[\frac{{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}\]

D) \[-\frac{{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}\]

Correct Answer: D

Solution :

             

Centripetal force on electron is

\[\frac{m{{v}^{2}}}{r}=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}\]

So, \[m{{v}^{2}}=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}\]

Kinetic energy \[=\frac{1}{2}m{{v}^{2}}=\frac{{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}\]

Potential energy \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{e}{r}(-e)=-\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}\]

Total energy = Kinetic energy+ Potential energy

\[\frac{{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}-\frac{{{e}^{2}}}{\pi {{\varepsilon }_{0}}r}=\frac{{{e}^{2}}-2{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}=\frac{-{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}\]