Respuesta :
Explanation:
It is known that [tex]mv_{r} = \frac{nh}{2p}[/tex]
where, m = mass of the electron
r = radius of the orbit
[tex]v_{r}[/tex] = orbital speed of the electron
Equation when the electron is experiencing uniform circular motion is as follows.
[tex]\frac{Kze^{2}}{r^{2}} = \frac{mv^{2}}{r}[/tex] ........ (1)
Rearranging above equation, we get the following.
[tex]mv^{2} = \frac{Kze^{2}}{r}[/tex]
Also, v = [tex]\frac{nh}{2pmr}[/tex] .......... (2)
Putting equation (2) in equation (1) we get the following.
[tex]\frac{mn^{2}h^{2}}{4p^{2}m^{2}r^{2}} = \frac{Kze^{2}}{r}[/tex]
Hence, formula for radius of the nth orbital is as follows.
[tex]r_{n} = [\frac{h^{2}}{4p^{2}mKze^{2}}]n^{2}[/tex]
[tex]r_{n} = [5.29 \times 10^{-11}m] \times (6)^{2}[/tex]
= [tex]19.044 \times 10^{-10} m[/tex]
= [tex]19.044 A^{o}[/tex]
Thus, we can conclude that the value for the radius r for a n= 6 Bohr orbit is [tex]19.044 A^{o}[/tex].
