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

Answer 1

Answer: (C)

In hydrogen atom electrostatic force of attraction $(F_e)$ between the revolving electrons and the nucleus provides the requisite centripetal force $(F_c)$ to keep them in their orbits. Thus,
$F_{e} = F_{c} $
$ \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}}$