Answer:
The diameter of the hydrogen [tex]\mathbf{d =1.0605 \times 10^{-10}\ m}[/tex]
Explanation:
From the given information:
Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:
[tex]L = \dfrac{nh}{2 \pi}[/tex]
Where the generic expression for angular momentum is:
L = mvr.
replacing the value of L into the previous equation, we have:
[tex]mvr= \dfrac{nh}{2 \pi}[/tex]
[tex]v= \dfrac{nh}{2 \pi mr}[/tex] ----- (1)
The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.
[tex]\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}[/tex] ----- (2)
replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:
[tex]\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2[/tex]
[tex]ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}[/tex]
[tex]r =\dfrac{n^2h^2}{4 \pi^2 mke^2}[/tex]
For ground-state n = 1
[tex]h = (6.625 \times 10^{-34} \ J.s)^2[/tex]
[tex]m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)[/tex]
[tex]Ke = (1.6 \times 10^{-19} \ C)^2[/tex]
[tex]r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}[/tex]
[tex]r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}[/tex]
[tex]\mathbf{r = 5.3025 \times 10^{-11} \ m}[/tex]
Therefore, the diameter of hydrogen d = 2r
[tex]\mathbf{d = ( 2 \times 5.3025 \times 10^{-11} \ m})}[/tex]
[tex]\mathbf{d =1.0605 \times 10^{-10}\ m}}[/tex]
