In the course of developing his model, Bohr arrived at the following formula for the radius of the electron’s orbit: rₙ = n²h²€₀/πmee², where m_{\mathrm{e}} is the electron mass, e is its charge, and ε₀ is a constant related to charge attraction in a vacuum. Given that me = 9.109×10⁻³¹ kg, e = 1.602×10⁻¹⁹C, €₀ = 80854×10⁻¹² C²/J . m , calculate the following:
(a) The radius of the first (n = 1) orbit in the H atom

The mean radius of an electron's orbit around the nucleus of a hydrogen atom in its ground state is known as the Bohr radius and is denoted by the letter a. (lowest-energy level). This radius' value, a, is determined by a physical constant that is roughly equivalent to 5.29177 x [tex]10^{ -11}[/tex] meters (m).

Substitute all values in the given formula

[tex]$\mathrm{r}_{n}=\frac{n^{2} h^{2} \varepsilon}{\pi m_{2} e^{2}}$[/tex]

[tex]$\mathrm{~h}=6.626^{*} 10^{-34} \mathrm{Js}$\\[/tex]

[tex]$\varepsilon=8.84510^{-12} \mathrm{C}^{2} / \mathrm{Jm}$[/tex]

[tex]$\mathrm{e}=1.602( 10^{-19} \mathrm{C})$[/tex]

[tex]$\mathrm{m}_{e}=9.109( 10^{-31} \mathrm{~kg})$[/tex]

[tex]$\mathrm{r}_{1}=\frac{\left(6.626 * 10^{-34}\right)^{2} (8.845 * 10^{-12})}{(3.14) ( 9.109 * 10^{-31}) \left(1.602 * 10^{-19}\right)^{2}}$[/tex]

Calculating, we get

[tex]$\mathrm{r}_{1}=5.29 $[/tex]×[tex]10^{-11} \mathrm{~m}[/tex]

This is the radius in the first orbit of the H atom.

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