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)

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 =√(4 π ε0/me2r) (B) v =√(e2/4 π ε0v) (C) v =√(e2/4 π ε0 mr) (D) v =√( v e2/4 π ε0 m )

Tardigrade Admin · Accepted Answer

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}

∴ \frac{m v ^{2}}{r} = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}}

or

v^{2} = \frac{e ^{2}}{4 π ε _{0} m r}

\Rightarrow v = \frac{e ^{2}}{4 π ε _{0} m r}

Q. 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)

$v = \frac{4 π ε _{0}}{m e ^{2} r}$

$v = \frac{e ^{2}}{4 π ε _{0} v}$

$v = \frac{e ^{2}}{4 π ε _{0} m r}$

$v = \frac{v e ^{2}}{4 π ε _{0} m}$

Q. 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)

v=me2r4πε0​​​

v=4πε0​ve2​​

v=4πε0​mre2​​

v=4πε0​mve2​​

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​ ∴rmv2​=4πε0​1​r2e2​ or v2=4πε0​mre2​ ⇒v=4πε0​mre2​​

$v = \frac{4 π ε _{0}}{m e ^{2} r}$

$v = \frac{e ^{2}}{4 π ε _{0} v}$

$v = \frac{e ^{2}}{4 π ε _{0} m r}$

$v = \frac{v e ^{2}}{4 π ε _{0} m}$