  Question

In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If $$a_0$$ is the radius of the ground state orbit, m is the mass and e is the charge on the election and $$\varepsilon_0$$ is the vacuum permittivity, the speed of the electron is?

A
0  B
eε0a0m  C
e4πε0a0m  D
4πε0a0me  Solution

The correct option is B $$\displaystyle\frac{e}{\sqrt{4\pi\varepsilon_0a_0m}}$$Centripetal force$$=\dfrac { m{ v }^{ 2 } }{ { a }_{ \circ } } \rightarrow (1)$$Colounbic force$$=\dfrac { 1 }{ 4\pi { \varepsilon }_{ \circ } } \dfrac { e.e }{ { a }^{ 2 } } \rightarrow (2)$$$$\therefore$$ Centripetal force= Coloumbic force$$\rightarrow \dfrac { m{ v }^{ 2 } }{ { a }_{ \circ } } \dfrac { 1 }{ 4\pi { \varepsilon }_{ \circ } } \dfrac { { e }^{ 2 } }{ { { a }_{ \circ } }^{ 2 } } \\ \rightarrow { v }^{ 2 }=\dfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ \circ }{ a }_{ \circ }m } \\ or\quad v=\dfrac { e }{ \sqrt { 4\pi { \varepsilon }_{ \circ }{ a }_{ \circ }m } }$$Physics

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