Question

The relation between the orbit radius and the electron velocity for a dynamically stable orbit in a hydrogen atom is (where, all notations have their usual meanings)

• A. $v = \sqrt{\frac{4\pi\varepsilon_{0}}{me^{2}r}}$
• B. $r = \sqrt{\frac{e^{2}}{4\pi\varepsilon_{0}v}}$
• C. $v = \sqrt{\frac{e^{2}}{4\pi\varepsilon_{0}mr}}$
• D. $r = \sqrt{\frac{ve^{2}}{4\pi\varepsilon_{0}m}}$

Answer 1

Answer: (C)

$v = \sqrt{\frac{e^{2}}{4\pi\varepsilon_{0}mr}}$

Answer 2

In hydrogen atom electrostatic force of attraction (Fe) between the revolving electrons and the nucleus provides the requisite centripetal force (fc) to keep them in their orbits. Thus,

Fe = Fc

$\therefore\frac{mv^{2}}{r} = \frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{r^{2}}$

Or $v^{2} = \frac{e^{2}}{4\pi\varepsilon_{0}mr} \Rightarrow v = \sqrt{\frac{e^{2}}{4\pi\varepsilon_{0}mr}}$